Introduction This chapter you will learn the SUVAT equations These are the foundations of many of the Mechanics topics You will see how to use them to

Introduction

• This chapter you will learn the SUVAT equations

• These are the foundations of many of the Mechanics topics

• You will see how to use them to use many types of problem involving motion

Kinematics of a Particle moving in a Straight Line

You will begin by learning two of the SUVAT equations

s = Displacement (distance)u = Starting (initial) velocityv = Final velocitya = Accelerationt = Time

2A

Replace with the appropriate letters.

Change in velocity = final velocity –

initial velocityMultiply by t

Add u

This is the usual form!

Replace with the

appropriate letters




2A

You need to consider using negative numbers in some cases

P Q

Positive direction

O4m 3m

2.5ms-1 6ms-1

If we are measuring displacements from O, and left to right is the positive direction…

For particle P: For particle Q:

The particle is to the left of the point O, which is the

negative direction

The particle is moving at 2.5ms-1 in the positive

direction

The particle is to the right of the

point O, which is the positive

direction

The particle is moving at 6ms-1 in the negative

direction




2A

A particle is moving in a straight line from A to B with constant acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to move from A to B. Find:a)The speed of the particle at Bb)The distance from A to B

A B

2ms-1

Start with a diagram

Write out ‘suvat’ and fill in what you

know

For part a) we need to calculate v, and we know u, a and

t…

Fill in the values you

know

Remember to include

units!

You always need to set up the question in this way. It makes it much easier to figure

out what equation you need to use (there will be more to learn than just these two!)




2A

A particle is moving in a straight line from A to B with constant acceleration 3ms-2. Its speed at A is 2ms-1 and it takes 8 seconds to move from A to B. Find:a)The speed of the particle at B – 26ms-1

b)The distance from A to B

A B

2ms-1

For part b) we need to calculate s, and we know u, v and

t…Fill in the

values you know

Show calculations

Remember the units!




2A

A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:a)The distance travelled over this 40 secondsb)The acceleration over the 40 seconds

4ms-1 7.5ms-1 Draw a diagram (model the cyclist

as a particle)


know

We are calculating s, and we already know u, v and t…Sub in the

values you know

Remember units!




2A

A cyclist is travelling along a straight road. She accelerates at a constant rate from a speed of 4ms-1 to a speed of 7.5ms-1 in 40 seconds. Find:a)The distance travelled over this 40 seconds – 230mb)The acceleration over the 40 seconds

4ms-1 7.5ms-1 Draw a diagram (model the cyclist

as a particle)


know

For part b, we are calculating a, and

we already know u, v and t…

Sub in the values you

know

Subtract 4

Divide by 40




2A

A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the speed 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

Draw a diagram


knowAs the particle is

decelerating, ‘a’ is negativeSub in the

values you know

Subtract 8

Divide by -1.5

A B




2A

A particle moves in a straight line from a point A to B with constant deceleration of 1.5ms-2. The speed of the particle at A is 8ms-1 and the speed 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

Draw a diagram


knowAs the particle is

decelerating, ‘a’ is negative

Sub in the values you

know

Calculate the answer!

A B




2A

After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find:a)The velocity of the particle at Cb)The distance from A to C

8ms-1 2ms-1

A B C

?

Update the diagram

Write out ‘suvat’ using

points A and C

Sub in the values

Work it out!

As the velocity is negative, this means the particle has now changed direction

and is heading back towards A! (velocity has a direction as well as a

magnitude!)The velocity is 1ms-1 in the direction C to

A…




2A

After reaching B the particle continues to move along the straight line with the same deceleration. The particle is at point C, 6 seconds after passing through A. Find:a)The velocity of the particle at C - -1ms-1

b)The distance from A to C

8ms-1 2ms-1

A B C

?

Update the diagram

Write out ‘suvat’ using

points A and C

Sub in the values

Work it out!

It is important to note that 21m is the distance from A to C only…

The particle was further away before it changed direction, and has in total travelled further than 21m…




2A

A car moves from traffic lights along a straight road with constant acceleration. The car starts from rest at the traffic lights and 30 seconds 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 kilometres per hourWe need to change the speed into metres per second first!

Draw a diagram

Multiply by 1000 (km to m)

Divide by 3600 (hours to seconds)




2A


Lights Trap

Draw a diagram

= 12.5ms-1


know

Sub in the values

Divide by 30

You can use exact

answers!




2A


Lights Trap

Draw a diagram

= 12.5ms-1


know

Sub in values

Work it out!


You can also use 3 more formulae linking different combination of

‘SUVAT’, for a particle moving in a straight line with constant

acceleration

2B

Subtract u

Divide by a

Replace t with the expression above

Multiply numerators and denominators

Multiply by 2a

Add u2

This is the way it is usually written!




acceleration

2B

Replace ‘v’ with ‘u + at’

Group terms on the numerator

Divide the numerator by 2

Multiply out the bracket




acceleration

2B

Subtract ‘at’

Replace ‘u’ with ‘v - at’ from above’

Multiply out the bracket

Group up the at2 terms


You can also use 3 more formulae linking different

combination of ‘SUVAT’, for a particle moving in a straight

line with constant acceleration

2B

A particle is moving in a straight line from A to B with constant acceleration 5ms-2. The velocity of the particle at A is 3ms-1 in the direction AB. The velocity at B is 18ms-1 in the same direction. Find the distance from A to B.

3ms-1 18ms-1

A B

Draw a diagram

Write out ‘suvat’ with the

information given

Replace v, u and a

Work out terms

Subtract 9

Divide by 10

We are calculating s,

using v, u and a





2B

A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a pillar 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

Pillar Box

Lamp Post

Draw a diagram


information given


using u, a and t

Replace u, a and t

Calculate





2B

A car is travelling along a straight horizontal road with a constant acceleration of 0.75ms-2. The car is travelling at 8ms-1 as it passes a pillar 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

Pillar Box

Lamp Post

Draw a diagram


information given

We are calculating v,

using u, a and t

Replace u, a and t

Calculate

Often you can use an answer you have calculated later on in the same question. However, you must

take care to use exact values and not rounded answers!





2B

A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with 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

Draw a diagram


information given

We are calculating t,

using s, u and aReplace s, u

and a

Simplify terms

Rearrange and set equal to 0

Factorise (or use the quadratic formula…)

We have 2 answers. As the acceleration is negative, the particle passes through A, then changes direction and

passes through it again!





2B

A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with 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

Draw a diagram


information given

We are calculating t,

using s, u and aThe particle passes through A at 2.5

seconds and 4 seconds, so it was beyond A for 1.5 seconds…





2B

A particle is moving in a straight horizontal line with constant deceleration 4ms-2. At time t = 0 the particle passes through a point O with 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

Draw a diagram


information given

The particle returns to O when s = 0

Replace s, u and a

Simplify

Rearrange

Factorise

The particle is at O when t = 0 seconds (to begin with)

and is at O again when t = 6.5 seconds





2B

A particle is travelling along the x-axis with constant deceleration 2.5ms-2. At time t = O, the particle passes through the origin, moving in the positive direction with speed 15ms-1. Calculate the distance travelled by the particle by the time it returns to the origin.

15ms-1

O X

Draw a diagram

The total distance travelled will be double the distance the particle reaches from O (point X)

At X, the velocity is 0

Replace v, u and a

Simplify

Add 5s

Divide by 5

45m is the distance from O to X. Double it for the total distance travelled


using u, v and a


You can use the formulae for constant acceleration to model an object moving vertically in a straight 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. As gravity 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 is another name for the object’s initial speed (u)

2C


You can use the formulae for constant acceleration to model an object moving vertically in a straight line under the influence

of gravity

2C

A ball is projected vertically upwards from a point O with a speed of 12ms-1. Find:a)The greatest height reached by the ballb)The total time the ball is in the air

12

ms-

10

ms-

1

Draw a diagram

At its highest point, the velocity of the ball is 0ms-1

As the ball has been projected upwards, gravity is acting in the opposite direction and hence the

acceleration is negative

Replace v, u and a

Simplify

Add 19.6s

Divide and round to 2sf (since gravity has been

given to 2sf)

We are calculating s, using u, v and a



of gravity

2C

A ball is projected vertically upwards from a point O with a speed of 12ms-1. Find:a)The greatest height reached by the ball – 7.4mb)The total time the ball is in the air

12

ms-

10

ms-

1

Draw a diagram

For the total time the ball is in the air, the

displacement (s) will be 0

Also, we will not know v (yet!) when the ball strikes

the ground

We are calculating t, using s, u and a

Replace s, u and a

Factorise

Choose the appropriate

answer!

So the ball will be in the air for 2.4 seconds



of gravity

2C

A book falls off the top shelf of a bookcase. The shelf is 1.4m above the ground. Find:a)The time it takes the book to reach the floorb)The speed with which the book strikes the floor

0m

s-1

Draw a diagram

1.4

m

The book’s initial speed will be 0 as it has not

been projected to begin with

As the book’s initial movement is downwards, we take the

acceleration due to gravity as positive

We are calculating t, using s, u and a…

Replace s, u and a

Simplify

Divide by 4.9

Find the positive square root



of gravity

2C

A book falls off the top shelf of a bookcase. The shelf is 1.4m above 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

0m

s-1

Draw a diagram

1.4

m

The book’s initial speed will be 0 as it has not

been projected to begin with

As the book’s initial movement is downwards, we take the

acceleration due to gravity as positive

We are calculating v, using s, u and a…

Replace s, u and a

Calculate




of gravity

2C

A ball is projected upwards from a point X which is 7m above the ground, with initial speed 21ms-1. Find the time of flight of the ball.

21

ms-

1

7m

Draw a diagram

The ball’s flight will last until it hits the ground

We want the ball to be 7m lower than it starts (in the

negative direction)Hence, s = -7

The ball is projected upwards, so the acceleration due to

gravity is negative


Replace s, u and a

Simplify

Rearrange and set equal to 0

We will need the quadratic formula here, so write down a, b and c…



of gravity

2C

A ball is projected upwards from a point X which is 7m above the ground, with initial speed 21ms-1. Find the time of flight of the ball.

21

ms-

1

7m

Draw a diagram

The ball’s flight will last until it hits the ground

We want the ball to be 7m lower than it starts (in the

negative direction)Hence, s = -7


gravity is negative

Replace a, b and c (using brackets!)

Calculate and be careful with any negatives in the

previous step!)



of gravity

2C

A particle is projected vertically upwards from a point O with initial speed u ms-1. The greatest height reached by the particle is 62.5m above the ground. Find:a)The speed of projectionb)The total time for which the ball is 50m or more above the ground

u m

s-1

62.5m Draw a diagram

The maximum height is 62.5m At this point the ball’s

velocity is 0ms-1


gravity is negative

We are calculating u, using s, v and a

Replace v, a and s

Simplify

Rewrite




of gravity

2C

A particle is projected vertically upwards from a point O with initial speed u ms-1. The greatest height reached by the particle is 62.5m above the ground. Find:a)The speed of projection – 35ms-1

b)The total time for which the ball is 50m or more above the ground

u m

s-1


The ball will pass the 50m mark twice – we need to find these

two times!

50m


Replace s, u and a

Simplify

Rearrange, and set equal to 0

We will need the quadratic formula, and hence a, b and

c



of gravity

2C

A particle is projected vertically upwards from a point O with initial speed u ms-1. The greatest height reached by the particle is 62.5m above the ground. Find:a)The speed of projection – 35ms-1

b)The total time for which the ball is 50m or more above the ground

u m

s-1


The ball will pass the 50m mark twice – we need to find these

two times!

50m


Sub these into the Quadratic formula

We get the two times the ball passes the 50m mark

Calculate the difference between these times!



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 projected vertically upwards from the bottom of the tower with velocity 21ms-1. The balls collide. Find the height at which this happens.

63

m

s1

s2

21ms-1

Draw a diagram

In this case we need to consider each ball separately.

We can call the two distances s1 and s2

The time will be the same for both when they collide, so we

can just use tMake sure that acceleration

is positive for A as it is travelling downwards and

negative for B as it is travelling upwards

Sub in s, u, a and t for

Ball B

Simplify

Sub in s, u, a and t for

Ball A

Simplify



of gravity

2C


63

m

s1

s2

21ms-1

Draw a diagram






negative for B as it is travelling upwards1)

2)Add the two equations together

(this cancels the 4.9t2 terms)

s1 + s2 must be the height of the tower (63m)

Divide by 21

So the balls collide after 3 seconds…



of gravity

2C


63

m

s1

s2

21ms-1

Draw a diagram






negative for B as it is travelling upwards2) Sub in t = 3 (we use

this equation since s2 is the height above the

ground)


You can represent the motion of an object 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!

v

u

t

On a speed-time graph, the Area beneath it is the

distance covered!


You can represent the motion of an object 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 = distance travelled 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, then decelerates uniformly for the final 20 seconds until it is at rest again.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 the rate of acceleration

jumps between different rates…

Time (s)

20 40 60 80 Time (s)

As the speed increases the curve gets steeper, but

with a constant speed the curve is straight. Finally the curve gets less steep

as deceleration takes place

Distance (m)


You can represent the motion of an object 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 = distance travelled

during that period

2D

The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms-1. She then decelerates at a constant rate, 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

8

12

6

Sub in the appropriate values for the trapezium

above

Calculate





period


during that period

2D

The diagram below shows a speed-time graph for the motion of a cyclist moving along a straight road for 12 seconds. For the first 8 seconds, she moves at a constant speed of 6ms-1. She then decelerates at a constant rate, 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

8 124

-6

Sub in the appropriate values for the trapezium

above

Calculate





period


during that period

2D

A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms-1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 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

5T

8

6T + 40

Sub in values

Simplify fraction

Divide by 8

Subtract 20

Divide by 5.5





period


during that period

2D

A particle moves along a straight line. It accelerates uniformly from rest to a speed of 8ms-1 in T seconds. The particle then travels at a constant speed for 5T seconds. It then decelerates to rest uniformly over the next 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 motionv(ms-1)

t(s)0

8

T 5T 405010

First section Last section

t(s)

a(ms-2)

20 40 60 80 100

0.8

-0.2





period


during that period

2D

A car C is moving along a straight road with constant speed 17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms-1 in 15 seconds and then 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

20

17.5

15

C

DAt the road sign, the cars have

covered the same distance in the same time

We need to set up simultaneous equations using s and t…

Let us call the time when the areas are equal ‘T’

T

17.5

T

T - 15

20

Sub in values

Sub in values

Simplify fraction

Multiply bracket





period


during that period

2D

A car C is moving along a straight road with constant speed 17.5ms-1. At time t = 0, C passes a lay-by. Also at time t = 0, a second car, D, leaves the lay-by. Car D accelerates from rest to a speed of 20ms-1 in 15 seconds and then 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

20

17.5

15

C

DAt the road sign, the cars have

covered the same distance in the same time

We need to set up simultaneous equations using s and t…

Let us call the time when the areas are equal ‘T’

T

Subtract 17.5T

Add 150Divide by 2.5

Sub in T

Calculate!

Set these equations equal to each other!

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 motion involving gravity

• We have also seen how to solve problems involving the motion of two particles

• We have also used graphs to solve some more complicated problems

Documents

Introduction This chapter you will learn the SUVAT equations These are the foundations of many of the Mechanics topics You will see how to use them to