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8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line
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8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line
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Introduction This chapter you will learn the SUVAT
equations
These are the foundations of many ofthe Mechanics topics
You will see how to use them to usemany types of problem involving motion
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8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line
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Kinematics of a Particle moving in aStraight Line
You will begin by learning two of theSUVAT equations
s = Displacement (distance)u = Starting (initial) velocity
v = Final velocitya = Accelerationt = Time
2A
=
=
=
+ =
= +
Replace with theappropriate letters. Change in velocity =
final velocity initialvelocity
Multiply by t
Add u
This is theusual form!
( ) =
= +2
Replacewith the
appropriateletters
8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line
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Kinematics of a Particle moving in aStraight Line
You will begin by learning two of theSUVAT equations
s = Displacement (distance)u = Starting (initial) velocity
v = Final velocitya = Accelerationt = Time
2A
= +
= + 2
You need to consider using negative numbers insome cases
P Q
Positive direction
O4m 3m
2.5ms -1 6ms -1
If we are measuring displacements from O, and left to rightis the positive direction
For particle P: For particle Q: = 4
= 2.5 = 3
= 6 The particle is to the left of
the point O, which is thenegative side
The particle is moving at2.5ms -1 in the positive direction
The particle is tothe right of thepoint O, which isthe positive side
The particle is moving at 6ms -1 in the negative direction
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A particle is moving in a straight line from A to B with constantacceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds tomove from A to B. Find:a) The speed of the particle at Bb) The distance from A to B
=? = 2 =? = 3 = 8 A B
2ms -1 Start with a
diagram
Write out suvat andfill in what you know
For part a) we needto calculate v, and we
know u, a and t = +
= 2 + (3 8)
= 26
Fill in thevalues you
know
Remember toinclude units!
You always need to set up the question in thisway. It makes it much easier to figure out whatequation you need to use (there will be more to
learn than just these two!)
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A particle is moving in a straight line from A to B with constantacceleration 3ms -2. Its speed at A is 2ms -1 and it takes 8 seconds tomove from A to B. Find:a) The speed of the particle at B 26ms -1b) The distance from A to B
=? = 2 =? = 3 = 8 A B
2ms -1
= 26
For part b) we needto calculate s, and we
know u, v and t =
+ 2
= 2 + 262
8
= 14 8
= 112
Fill in thevalues you
knowShow
calculations
Rememberthe units!
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A cyclist is travelling along a straight road. She accelerates at aconstant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40seconds. Find:a) The distance travelled over this 40 secondsb) The acceleration over the 40 seconds
4ms -1 7.5ms -1
=? = 4 =? = 40 = 7.5
Draw a diagram(model the cyclist as
a particle)
Write out suvat andfill in what you know
= + 2
We are calculating s,and we already know
u, v and t
= 4 + 7.52
40
= 230
Sub in thevalues you
know
Rememberunits!
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A cyclist is travelling along a straight road. She accelerates at aconstant rate from a speed of 4ms -1 to a speed of 7.5ms -1 in 40seconds. Find:a) The distance travelled over this 40 seconds 230mb) The acceleration over the 40 seconds
4ms -1 7.5ms -1
= 230 = 4 =? = 40 = 7.5
Draw a diagram(model the cyclist as
a particle)
Write out suvat andfill in what you know
For part b, we arecalculating a, and wealready know u, v and
t
Sub in thevalues you
know
Subtract 4
= +
7.5 = 4 + 40
7.5 = 40 Divide by
40 = 0.0875
=?
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A particle moves in a straight line from a point A to B with constantdeceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and thespeed of the particle at B is 2ms -1. Find:a) The time taken for the particle to get from A to Bb) The distance from A to B
8ms -1 2ms -1
=? = 8 = 1.5 =? = 2
Draw a diagram
Write out suvat andfill in what you know
As the particle isdecelerating, a is
negative = +
2 = 8 1.5
6 = 1.5
4 =
Sub in thevalues you know
Subtract 8
Divide by -1.5
A B
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A particle moves in a straight line from a point A to B with constantdeceleration of 1.5ms -2. The speed of the particle at A is 8ms -1 and thespeed of the particle at B is 2ms -1. Find:a) The time taken for the particle to get from A to B 4 secondsb) The distance from A to B
8ms -1 2ms -1
=? = 8 = 1.5 =? = 2
Draw a diagram
Write out suvat andfill in what you know
As the particle isdecelerating, a is
negativeSub in the
values you know
= 4
= + 2
= 8 + 22
4
= 20
Calculate theanswer!
A B
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
After reaching B the particle continues to move along the straight linewith the same deceleration. The particle is at point C, 6 seconds afterpassing through A. Find:a) The velocity of the particle at Cb) The distance from A to C
8ms -1 2ms -1
=? = 8 = 1.5 =? = 6 A B C
?Update the
diagram
Write outsuvat using
points A and C
= +
= 8 (1.5 6)
= 1
Sub in thevalues
Work itout!
As the velocity is negative, this means theparticle has now changed direction and isheading back towards A! (velocity has a
direction as well as a magnitude!)
The velocity is 1ms -1 in the direction C to A
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
After reaching B the particle continues to move along the straight linewith the same deceleration. The particle is at point C, 6 seconds afterpassing through A. Find:a) The velocity of the particle at C - -1ms-1b) The distance from A to C
8ms -1 2ms -1
=? = 8 = 1.5 =? = 6 A B C
?Update the
diagram
Write outsuvat using
points A and C
= 1
= + 2
= 8 12
6
= 21
Sub in thevalues
Work itout!
It is important to note that 21m is the distance from A to Conly
The particle was further away before it changeddirection, and has in total travelled further than 21m
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A car moves from traffic lights along a straight road with constantacceleration. The car starts from rest at the traffic lights and 30seconds later passes a speed trap where it is travelling at 45 kmh -1. Find:a) The acceleration of the carb) The distance between the traffic lights and the speed-trap.
0ms -1 45kmh -1
Lights Trap
Standard units to use are metres and seconds, or kilometres and hoursIn this case, the time is in seconds and the speed is in kilometresper hourWe need to change the speed into metres per second first!
Draw a diagram
45
45,000
12.5
Multiply by 1000 (km to m)
Divide by 3600 (hours to seconds)
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A car moves from traffic lights along a straight road with constantacceleration. The car starts from rest at the traffic lights and 30seconds later passes a speed trap where it is travelling at 45 kmh -1. Find:a) The acceleration of the carb) The distance between the traffic lights and the speed-trap.
0ms -1 45kmh -1
Lights Trap
Draw a diagram
= 12.5ms-1
Write out suvat andfill in what you know =?
= 0 =? = 30 = 12.5
= +
12.5 = 0 + 30
512
=
Sub in thevalues
Divide by30
You can useexact answers!
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Kinematics of a Particle moving in aStraight Line
You will begin by learningtwo of the SUVAT
equations
s = Displacement (distance)
u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time
2A
= +
= +2
A car moves from traffic lights along a straight road with constantacceleration. The car starts from rest at the traffic lights and 30seconds later passes a speed trap where it is travelling at 45 kmh -1. Find:a) The acceleration of the carb) The distance between the traffic lights and the speed-trap.
0ms -1 45kmh -1
Lights Trap
Draw a diagram
= 12.5ms-1
Write out suvat andfill in what you know =?
= 0 =? = 30 = 12.5 =5
12
= + 2
=0 + 12.5
2 30
= 187.5
Sub in
values
Work itout!
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8/12/2019 2) M1 Kinematics of a Particle Moving in a Straight Line
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulae linkingdifferent combination of SUVAT, for a
particle moving in a straight line withconstant acceleration
2B
= +
= + 2
= +
=
=
= + 2
= + 2
= 2
2 =
+ 2 =
= + 2
= + 2
Subtract u
Divide by a
Replace t with theexpression above
Multiply numerators anddenominators
Multiply by 2a
Add u2
This is the way it isusually written!
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulae linkingdifferent combination of SUVAT, for a
particle moving in a straight line withconstant acceleration
2B
= +
= + 2
= + 2
= + 2
= + +
2
=2 +
2
= +12
= +12
Replace v with u + at
Group terms on thenumerator
Divide the numeratorby 2
Multiply out thebracket
= +1
2
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulae linkingdifferent combination of SUVAT, for a
particle moving in a straight line withconstant acceleration
2B
= +
= + 2
= + 2
= +1
2
= +
=
= +12
= ( ) +12
= +12
= 12
= 12
Subtract at
Replace u with v- atfrom above
Multiply out thebracket
Group up the at 2 terms
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A particle is moving in a straight line from A to B with constantacceleration 5ms -2. The velocity of the particle at A is 3ms -1 in thedirection AB. The velocity at B is 18ms -1 in the same direction. Find thedistance from A to B.
3ms -1 18ms-1
A B
Draw a diagram
=? = 3 = 5 =? = 18 Write out suvat
with theinformation given
= + 2
18 = 3 + 2(5)
324 = 9 + 10
315 = 10
31.5 =
Replace v, u and a
Work out terms
Subtract 9
Divide by 10
We arecalculating s,
using v, u and a
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A car is travelling along a straight horizontal road with a constantacceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes apillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp postb) The speed with which the car passes the lamp post
8ms -1
PillarBox
LampPost
=? = 8 = 0.75 = 12 =?
Draw a diagram
Write out suvat with the
information given
We arecalculating s,
using u, a and t = +12
= (8 12) +12
(0.75 12 )
= 150
Replace u, aand t
Calculate
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A car is travelling along a straight horizontal road with a constantacceleration of 0.75ms -2. The car is travelling at 8ms -1 as it passes apillar box. 12 seconds later the car passes a lamp post. Find:a) The distance between the pillar box and the lamp post 150mb) The speed with which the car passes the lamp post
8ms -1
PillarBox
LampPost
=? = 8 = 0.75 = 12 =?
Draw a diagram
Write out suvat with the
information given
We arecalculating v,
using u, a and tReplace u, a
and t
Calculate
= +
= 8 + (0.75 12)
= 17
Often you can use an answer you have calculated lateron in the same question. However, you must take care
to use exact values and not rounded answers!
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through Ab) The total time the particle is beyond Ac) The time taken for the particle to return to O
13ms-1
O A = 20 = 13 = 4 =? =?
Draw a diagram
Write out suvat with the
information given
We arecalculating t,
using s, u and a = +12
20 = (13) + 12
( 4)
20 = 13 2
2 13 + 20 = 0
(2 5)( 4) = 0
= 2.5 4
Replace s, u
and a
Simplify terms
Rearrange and set equal to 0
Factorise (or use the quadratic formula)
We have 2 answers. As theacceleration is negative, the
particle passes through A, thenchanges direction and passes
through it again!
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A 2.5 and 4 secondsb) The total time the particle is beyond Ac) The time taken for the particle to return to O
13ms-1
O A = 20 = 13 = 4 =? =?
Draw a diagram
Write out suvat with the
information given
We arecalculating t,
using s, u and aThe particle passes through A at 2.5
seconds and 4 seconds, so it wasbeyond A for 1.5 seconds
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A particle is moving in a straight horizontal line with constantdeceleration 4ms -2. At time t = 0 the particle passes through a point Owith speed 13ms -1, travelling to a point A where OA = 20m. Find:a) The times when the particle passes through A 2.5 and 4 secondsb) The total time the particle is beyond A 1.5 secondsc) The time taken for the particle to return to O
13ms-1
O A = 20 = 13 = 4 =? =?
Draw a diagram
Write out suvat with the
information given
The particlereturns to Owhen s = 0
= 0
= +12
0 = (13) + ( 2)
0 = 13 2
2 13 = 0
(2 13) = 0
Replace s, u and a
= 0 6.5
Simplify
Rearrange
FactoriseThe particle is at O when t = 0seconds (to begin with) and is
at O again when t = 6.5 seconds
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Kinematics of a Particle moving in aStraight Line
You can also use 3 more formulaelinking different combination of
SUVAT, for a particle moving ina straight line with constant
acceleration
2B
= + = +
2
= + 2 = +12
= 12
A particle is travelling along the x-axis with constant deceleration2.5ms -2. At time t = O, the particle passes through the origin, movingin the positive direction with speed 15ms -1. Calculate the distancetravelled by the particle by the time it returns to the origin.
15ms-1
O X
Draw a diagram
The total distancetravelled will be double the
distance the particlereaches from O (point X)
At X, the velocity is 0
=? = 15 = 2.5 =? = 0
= + 2
0 = 15 + 2( 2.5)
0 = 225 5
5 = 225
= 45
Replace v,u and a
Simplify
Add 5s
Divide by 5
= 90
45m is the distance from Oto X. Double it for the total
distance travelled
We arecalculating s,
using u, v and a
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constant acceleration to model an object moving vertically in astraight line under the influence of gravity
Gravity causes objects to fall to the earth! (as you probably already know!)
The acceleration caused by gravity is constant (if you ignore air resistance)
This means the acceleration will be the same, regardless of the size of the object
On Earth, the acceleration due to gravity is 9.8ms -2 , correct to 2 significant figures.
When solving problems involving vertical motion you must carefully consider the direction. Asgravity acts in a downwards direction:
- An object thrown downwards will have an acceleration of 9.8ms -2- An object thrown upwards will have an acceleration of -9.8ms -2
The time of flight is the length of time an object spends in the air. The speed of projection isanother name for the objects initial speed (u)
2C
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball is projected vertically upwards from a point O with a speedof 12ms -1. Find:a) The greatest height reached by the ballb) The total time the ball is in the air
= + = + 2
= + 2 = +12
= 12
1 2 m s -
1
0 m s -
1
=?
= 12
= 9.8
=?
= 0
Draw a diagram
At its highest point, thevelocity of the ball is 0ms -1
As the ball has been projectedupwards, gravity is acting in theopposite direction and hence the
acceleration is negative
0 = 12 + 2( 9.8)
= + 2
0 = 144 19.6
19.6 = 144
= 7.4 (2 )
Replace v, u and a
Simplify
Add 19.6s
Divide and round to 2sf (sincegravity has been given to 2sf)
We are calculating s, using u, v and a
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball is projected vertically upwards from a point O with a speedof 12ms -1. Find:a) The greatest height reached by the ball 7.4mb) The total time the ball is in the air
= + = + 2
= + 2 = +12
= 12
1 2 m s -
1
0 m s -
1
= 12
= 9.8
=?
Draw a diagram
For the total time the ball isin the air, the displacement
(s) will be 0
= 0
=?
Also, we will not know v (yet!)when the ball strikes the
ground
We are calculating t, using s, u and a
= + 12
0 = 12 4.9
0 = (12 4.9 )
= 0 2.4 (2 )
Replace s, u and a
Factorise
Choose theappropriate answer!
So the ball will be in the air for 2.4 seconds
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A book falls off the top shelf of a bookcase. The shelf is 1.4above the ground. Find:a) The time it takes the book to reach the floorb) The speed with which the book strikes the floor
= + = + 2
= + 2 = +12
= 12
0 m s -1
= 0
= 9.8
=?
= 1.4
=?
Draw a diagram
1 . 4 m
The books initial speed willbe 0 as it has not been
projected to begin withAs the books initial movement is
downwards, we take theacceleration due to gravity as
positive
We are calculating t, usings, u and a = +
1
2
1.4 = (0) +12
(9.8)
1.4 = 4.9
1.44.9
=
= 0.53 (2 )
Replace s, u and a
Simplify
Divide by 4.9
Find the positivesquare root
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A book falls off the top shelf of a bookcase. The shelf is 1.4above the ground. Find:a) The time it takes the book to reach the floor 0.53 secondsb) The speed with which the book strikes the floor
= + = + 2
= + 2 = +12
= 12
0 m s -1
= 0
= 9.8
=?
= 1.4
=?
Draw a diagram
1 . 4 m
The books initial speed willbe 0 as it has not been
projected to begin withAs the books initial movement is
downwards, we take theacceleration due to gravity as
positive
We are calculating v, usings, u and a
= + 2
= 0 + 2(9.8 1.4)
= 27.44
= 5.2 (2 )
Replace s, uand a
Calculate
Find the positivesquare root
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball is projected upwards from a point X which is 7m above theground, with initial speed 21ms -1. Find the time of flight of theball.
= + = + 2
= + 2 = +12
= 12
2 1 m s -
1
7 m
= 21
= 9.8
=?
= 7
=?
Draw a diagram
The balls flight will last until ithits the ground
We want the ball to be 7mlower than it starts (in the
negative direction)Hence, s = -7
The ball is projected upwards, sothe acceleration due to gravity is
negative
= +12
7 = (21) +12 ( 9.8)
7 = 21 4.9
4.9 21 7 = 0
= 4.9 = 21 = 7
We are calculating t,using s, u and aReplace s, uand a
Simplify
Rearrange and set equal to 0
We will need the quadratic formulahere, so write down a, b and c
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball is projected upwards from a point X which is 7m above theground, with initial speed 21ms -1. Find the time of flight of theball.
= + = + 2
= + 2 = +12
= 12
2 1 m s -
1
7 m
= 21
= 9.8
=?
= 7
=?
Draw a diagram
The balls flight will last until ithits the ground
We want the ball to be 7mlower than it starts (in the
negative direction)Hence, s = -7
The ball is projected upwards, sothe acceleration due to gravity is
negative = 4.9 = 21 = 7
= 4
2
=( 21) ( 21) (4 4.9 7)
(2 4.9)
= 4.6 0.3
Replace a, b and c(using brackets!)
Calculate and be careful withany negatives in the previous
step!)
= 4.6 (2 )
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projectionb) The total time for which the ball is 50m or more above the ground
= + = + 2
= + 2 = +12
= 12
u m s -
1
62.5m Draw a diagram
The maximum height is 62.5m
At this point the balls velocityis 0ms -1
The ball is projected upwards, sothe acceleration due to gravity is
negative
=?
= 9.8
=?
= 62.5
= 0
We are calculating u,using s, v and a
= + 2
0 = + 2( 9.8 62.5)
0 = 1225
= 1225
= 35
Replace v, a and s
Simplify
Rewrite
Find the positive square root
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projection 35ms -1b) The total time for which the ball is 50m or more above the ground
= + = + 2
= + 2 = +12
= 12
u m s -
1
62.5m Draw a diagram
The ball will pass the 50m marktwice we need to find these two
times!
= 9.8
=?
= 50
=?
= 35 50m
We are calculating t,using s, u and a
= +
1
2
50 = (35) +12
( 9.8)
50 = 35 4.9
4.9 35 + 50 = 0
= 4.9 = 35 = 50
Replace s, u and a
Simplify
Rearrange, and set equal to 0
We will need the quadratic
formula, and hence a, b and c
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A particle is projected vertically upwards from a point O with initialspeed u ms-1. The greatest height reached by the particle is 62.5mabove the ground. Find:a) The speed of projection 35ms -1b) The total time for which the ball is 50m or more above the ground
= + = + 2
= + 2 = +12
= 12
u m s -
1
62.5m Draw a diagram
The ball will pass the 50m marktwice we need to find these two
times!
= 9.8
=?
= 50
=?
= 35 50m
We are calculating t,using s, u and a
= 4.9 = 35 = 50
Sub these into theQuadratic formula
= 4
2
=( 35) ( 35) (4 4.9 50)
4.9 2
= 5.1686 = 1.9742
We get the two times the ballpasses the 50m mark
Calculate the difference
between these times! = 3.2 (2 )
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball, A, falls vertically from rest from the top of a tower 63m high.At the same time as A begins to fall, another ball, B, is projectedvertically upwards from the bottom of the tower with velocity 21ms -1.The balls collide. Find the height at which this happens.
= + = + 2
= + 2 = +12
= 12
6 3 m
s1
s221ms-1
Draw a diagram
In this case we need to considereach ball separately.
We can call the two distancess1 and s 2
The time will be the same forboth when they collide, so we
can just use tMake sure that acceleration
is positive for A as it istravelling downwards and
negative for B as it istravelling upwards
= 9.8 =
=
=?
= 0
= 9.8 =
=
=?
= 21
= + 12
= (0) +12
(9.8)
= 4.9
= + 12
= (21) +12
( 9.8)
= 21 4.9
Sub in s, u,a and t for
Ball B
Simplify
Sub in s, u,a and t for
Ball A
Simplify
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball, A, falls vertically from rest from the top of a tower 63m high.At the same time as A begins to fall, another ball, B, is projectedvertically upwards from the bottom of the tower with velocity 21ms -1.The balls collide. Find the height at which this happens.
= + = + 2
= + 2 = +12
= 12
6 3 m
s1
s221ms-1
Draw a diagram
In this case we need to considereach ball separately.
We can call the two distancess1 and s 2
The time will be the same forboth when they collide, so we
can just use tMake sure that acceleration
is positive for A as it istravelling downwards and
negative for B as it istravelling upwards
= 9.8 =
=
=?
= 0
= 9.8 =
=
=?
= 21
= 4.9
= 21 4.9
1)
2) Add the two equations together(this cancels the 4.9t 2 terms)
+ = 21
63 = 21
3 =
s1 + s2 must be the height of thetower (63m)
Divide by 21
So the balls collide
after 3 seconds
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Kinematics of a Particle moving in aStraight Line
You can use the formulae for constantacceleration to model an object movingvertically in a straight line under the
influence of gravity
2C
A ball, A, falls vertically from rest from the top of a tower 63m high.At the same time as A begins to fall, another ball, B, is projectedvertically upwards from the bottom of the tower with velocity 21ms -1.The balls collide. Find the height at which this happens.
= + = + 2
= + 2 = +12
= 12
6 3 m
s1
s221ms-1
Draw a diagram
In this case we need to considereach ball separately.
We can call the two distancess1 and s 2
The time will be the same forboth when they collide, so we
can just use tMake sure that acceleration
is positive for A as it istravelling downwards and
negative for B as it istravelling upwards
= 9.8 =
=
=?
= 0
= 9.8 =
=
=?
= 21
= 21 4.9 2) Sub in t = 3 (we use thisequation since s
2 is the
height above the ground) = 21(3) 4.9(3)
= 18.9 (19 2 )
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Kinematics of a Particle moving in aStraight Line
You can represent the motion of anobject on a speed-time graph,
distance-time graph or an acceleration-time graph
2D
O
u
v
t
Initial velocity
Final velocity
Time taken
v - u
t
=
=
=
On a speed-time graph,the gradient of a section
is its acceleration!
= + 2
=
= +
2
v
u
t
= + 2
On a speed-time graph,the Area beneath it isthe distance covered!
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Kinematics of a Particle moving in aStraight Line
You can represent the motion of anobject on a speed-time graph,
distance-time graph or an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph = distancetravelled during that period
2D
A car accelerates uniformly at 5ms -2 from rest for 20 seconds.It then travels at a constant speed for the next 40 seconds, thendecelerates uniformly for the final 20 seconds until it is at restagain.a) Draw an acceleration-time graph for this informationb) Draw a distance-time graph for this information
20 40 60 80
5
Acceleration(ms-2)
0
-5
For now, we assume therate of acceleration jumps between different
rates Time (s)
20 40 60 80Time (s)
As the speed increases thecurve gets steeper, but witha constant speed the curve is
straight. Finally the curvegets less steep as
deceleration takes place
Distance(m)
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
The diagram below shows a speed-time graph for the motion of a cyclistmoving along a straight road for 12 seconds. For the first 8 seconds, shemoves at a constant speed of 6ms -1. She then decelerates at a constantrate, stopping after a further 4 seconds. Find:a) The distance travelled by the cyclistb) The rate of deceleration of the cyclist
v(ms-1)
t(s)0
6
8 12
= + 2
=8 + 12
2 6
8
12
6
= 60
60
Sub in the appropriate valuesfor the trapezium above
Calculate
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
The diagram below shows a speed-time graph for the motion of a cyclistmoving along a straight road for 12 seconds. For the first 8 seconds, shemoves at a constant speed of 6ms -1. She then decelerates at a constantrate, stopping after a further 4 seconds. Find:a) The distance travelled by the cyclist 60mb) The rate of deceleration of the cyclist
v(ms-1)
t(s)0
6
8 12
=
4
-6
Sub in the appropriate valuesfor the trapezium above
Calculate =
64
= 1.5
1.5
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
A particle moves along a straight line. It accelerates uniformly from restto a speed of 8ms -1 in T seconds. The particle then travels at a constantspeed for 5T seconds. It then decelerates to rest uniformly over thenext 40 seconds.a) Sketch a speed-time graph for this motionb) Given that the particle travels 600m, find the value of Tc) Sketch an acceleration-time graph for this motion
v(ms-1)
t(s)0
8
T 5T 40
= + 2
600 =5 + 6 + 40
2 8
5T
8
6T + 40
600 = 5.5 + 20 8
75 = 5.5 + 20
55 = 5.5
10 =
Sub invalues
Simplifyfraction
Divideby 8
Subtract 20
Divide by 5.5
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
A particle moves along a straight line. It accelerates uniformly from restto a speed of 8ms -1 in T seconds. The particle then travels at a constantspeed for 5T seconds. It then decelerates to rest uniformly over thenext 40 seconds.a) Sketch a speed-time graph for this motionb) Given that the particle travels 600m, find the value of T 10 secondsc) Sketch an acceleration-time graph for this motion
v(ms-1)
t(s)0
8
T 5T 405010
=
=8
10
= 0.8
=8
40
= 0.2
First section Last section
t(s)
a(ms-2)
20 40 60 80 100
0.8
-0.2
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
A car C is moving along a straight road with constant speed 17.5ms -1. Attime t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves thelay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds andthen maintains this speed. Car D passes car C at a road sign.a) Sketch a speed-time graph to show the motion of both carsb) Calculate the distance between the lay-by and the road sign
v(ms-1)
t(s)0
2017.5
15
CD
At the road sign, the cars have
covered the same distance in thesame time
We need to set up simultaneousequations using s and t
Let us call the time when theareas are equal T
T
17.5
T
=
= 17.5
= 17.5
= + 2
= + 15
2 20
T - 15
20
= 7.5 20
= 20 150
Sub invalues
Sub invalues
Simplifyfraction
Multiply
bracket
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Kinematics of a Particle moving in aStraight Line
You can represent the motionof an object on a speed-timegraph, distance-time graph or
an acceleration-time graph
Gradient of a speed-time graph =Acceleration over that period
Area under a speed-time graph =distance travelled during that
period
2D
A car C is moving along a straight road with constant speed 17.5ms -1. Attime t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves thelay-by. Car D accelerates from rest to a speed of 20ms -1 in 15 seconds andthen maintains this speed. Car D passes car C at a road sign.a) Sketch a speed-time graph to show the motion of both carsb) Calculate the distance between the lay-by and the road sign
v(ms-1)
t(s)0
2017.5
15
CD
At the road sign, the cars have
covered the same distance in thesame time
We need to set up simultaneousequations using s and t
Let us call the time when theareas are equal T
T
= 17.5
= 20 150
17.5 = 20 150
0 = 2.5 150
150 = 2.5
60 =
= 17.5
= 17.5(60)
= 1050
Subtract17.5T
Add150Divideby 2.5
Subin T
Calculate!
Set theseequations equal to
each other!
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Summary This chapter we have seen how to solve problems
involving the motion of a particle in a straight line,with constant acceleration
We have extended the problems to vertical motioninvolving gravity
We have also seen how to solve problems involvingthe motion of two particles
We have also used graphs to solve some morecomplicated problems