AQA Physics Suvat equations

SUVAT equations of motion Revision

questions & answers

A 'Level Physics

AQA

by Y Malik

What you need to know

• Speed and velocity

• Distance – time graphs

• Velocity

• Displacement-time graphs

• Acceleration

• Free fall

• Motion graphs • (projectile motion is covered in a separate lecture)

by y.malik

Equations you should know

Speed and velocity

Speed v = s/t

v = 2πr/t

Average speed = s/t

v = Δs/Δt

g= 9.8ms2

Acceleration

a= (v-u)/t

v= u+at

s= ½ (u+v)t

s= ut+ ½ at2

v2=u2+2as

by y.malik

Suvat equations of motion

• S= displacement (m)

• U = initial velocity (ms-1)

• V= final velocity (ms-1)

• A= acceleration (ms-2)

• T = time (s)

v=u+at this is a rearrangement of a =(v-u)/t

s= ½ (v+u)t this says displacement = average velocity x time

s= ut+ ½ at2 with zero acceleration, this becomes displacement

= velocity x time

V2+u2+2as useful when you don’t know the time

by y.malik

Free fall and the acceleration of gravity. 9.8ms-2

• Free falling object has an acceleration of 9.8ms-2 downward on Earth

• Known as acceleration gravity

• If the velocity and time for a free falling object being dropped from a position of rest where tabulated, then the following pattern would emerge:

• time (s) velocity (m/s) 0 0

1 -9.8

2 -19.6

3 -29.4

4 -39.2

5 -49.0

by y.malik

Free fall and the acceleration of gravity. 9.8ms-2

• You would observe that the velocity time data reveals that the object is changing by 9.8 ms-1 each consecutive second

• Therefore we say the freefalling object has an acceleration of approximately 9.8ms-2 (that’s 9.8 meters per second per second)

by y.malik

Significant figures

• Calculator displays large number of digits

• Always round up or down the final answer of a calculation to the same number of significant figures as the data given

• Sometimes an answer to one part of a question has to be used in a subsequent calculation, in which case, the numerical answer to the first part should be carried forward without rounding it up or down

Ref: AQA

by y.malik

Q:An aeroplane taking off accelerates uniformly on a runway from a velocity of 4ms-1 to a velocity of 64ms-1 in 40s. Calculate its acceleration.

A: 1.5ms-2

Method:

S= To find acceleration we use

U= 4ms-1 a= (v-u)/t

V= 64ms-1 = (64-4)/40s

A= ? = 1.5ms-2

T= 40s

by y.malik

Q: A car travelling a velocity of 20ms-1 brakes to a standstill in 8.0s. Calculate its deceleration, assuming its velocity decreases uniformly.

A: -2.5ms-2

Method:

S= To find acceleration we use

U= 4ms-1 a= (v-u)/t

V= 0ms-1 = (20-0)/8s

A=? = -2.5ms-2

T= 8.0s

by y.malik

Q: Cyclist accelerates uniformly from a velocity of 2.5ms-1 to a velocity of 7.0ms-1 in a time of 10s. Calculate a) its acceleration, b) its velocity 2.0s later if it continued to accelerate at the same rate.

A: a) 0.45ms-2 b) 7.9ms-1 Method: S= U= 2.5ms-1 a) a= (v-u)/t = 0.45ms^^ V= 7.0ms-1 b) v=u+at A= ? = 2.5 + (0.45x12s) T= 10s +2s = 7.9 ms-1 by y.malik

Q: A driver of a vehicle travelling at a speed of 30ms-1 on a motorway brakes to a standstill in a distance of 100m. Calculate the deceleration of the vehicle.

A: -4.5ms-2

Method:

S= 100m we have s, u, v, and need, a

U= 30ms-1 we use, v2=u2+2as

V= 0ms-1 0=302+(2xax100)

A= ? a = 900/200

T= a = 4.5ms-2

by y.malik

Q: A pebble released at rest from a canal bridge, took 0.9s to hit the water. Calculate a) distance it fell before hitting the water, b)its speed just before hitting the water

A: a) -12.5m b) -15.7ms-1

Method: a) to find distance, use:

S= ? S= ½ at2

U= 0ms-1 = ½ x -9.8 x 1.62

V= ? = -12.5 m

A= 9.8ms-1 b) to find v, use:

T= 0.9s v=u+at = -15.7ms-1

by y.malik

Q: A ball was dropped from a hot air balloon when the balloon was at rest 50m above the ground. Calculate: a) the time taken for the ball to hit the ground, b) the speed on hitting the ground

A: a) 3.2s b) 31 ms-1

S= 50m a) s= ½ at2 b) v= u+at

U= 0 50m= ½ x -9.8 x t2 = 0+9.8x3.2

V= ? 50m = 4.9 x t2 = 31ms-1

A= t2 = 50m / 4.9

T= ? t = √ 10.20408163

t = 3.19 = 3.2s

by y.malik

Q: An astronaut on the Moon threw an object 4.0m vertically upwards and caught it again 4.5s later. Calculate: a) its acceleration due to gravity on the moon b) the speed of projection of the object c) how high the object would have risen on the Earth for the same speed of projection

a) 1.6 ms-2 moon is ¼ the size of Earth. So the moons gravity is much less than the Earths gravity. The acceleration in freefall on the moon is about 1.622 m2 (approx 0.165 Earth gravity)

by y.malik

b) V2=u2+2as

= 02+ (2x1.6x4.0)

= √12.8

= 3.577708764

= 3.6 ms-2

c)

S= ? v2=2as

U= 0 = 2x9.8x s

V= 3.6 s= 12.96/19.6 = 0.65m

A=9.8

T=4.5

by y.malik

Q: A swimmer swims 100m from one end of a swimming pool to the other end at a constant speed of 1.2 ms-1 then swims back at constant speed returning to the starting point 210s after starting.

Calculate how long the swimmer takes to swim from a) the starting end to the other end, b) back to the start

from the other end

A: a) t=s/v = 100m/1.2ms-1 = 83s

b) 210s – 83s = 127s

by y.malik

Q: A ball released from a height of 1.8m above a level surface and rebounds to a height of 0.90m

a) Given that g=9.8ms-1 calculate i) the duration of its descent, ii) its velocity just before impact, iii) the duration of its ascent, iv) its velocity just after impact

b) Sketch a graph to show how its velocity changes with time from release to rebound at maximum height

c) Sketch a graph to show how the displacement of the object changes with time

by y.malik

ai) you are looking for the time.

S=1.8m s = ut+ ½ at2

U=0 = (0xt=0) + ½ x 9.8 x t2

V= 1.8 = ½ x 9.8 x t2

A=9.8 t = √ (1.8 / 4.9)

T= ? t= 0.61s (duration of its fall downwards)

aii)

S=1.8m you can use either equations v=u+at or v2 = u2 +2as

U=0 = 0+(9.8x0.61) = 2x9.8x18

V=? = 5.9 ms-1 = √35.28

A=9.8 (this is velocity B4 impact = 5.9 ms-1

T=0.61

by y.malik

aiii) duration of its ascent

S=0.90 s = ut+ ½ at2

U=5.9 = (0xt=0) + ½ x 9.8 x t2

V=0 0.9 = ½ x 9.8 x t2

A=9.8 t = √ 0.1836

T= ? t= 0.43s (this is the rebound)

aiv) velocity just after impact

S=0.9m =this is the rebound distance v2 = u2 +2as

U=0 =at impact the initial velocity is zero = 2x9.8x0.9

V=? = Rebound final velocity = √17.64

A=9.8 = 4.2ms-1

T=0.43 (rebound upwards velocity)

by y.malik

b) diagram to show velocity changes with time from release to rebound at maximum height

4.2ms-1

-5.9ms-1

0

Velocity ms-1

time (s)

Ball released so falls downwards

Time taken for downwards fall

0.61s

Ball rebounds upwards

Time taken for rebound

1.04

0.61+0.43=1.04s

Ball released. Falls for 0.61s with velocity of -5.9ms^ B4 impact Then rebounds upwards (ascent) for 0.43s at velocity of 4.2ms^

by y.malik

c) Graph to show how displacement of the object changes with time

Max height

1.8m

displacement (m)

0.61s t(s)

1.04

by y.malik

Q: A car on a straight downhill road accelerates uniformly from 4.0ms-1 to 29ms-1 over a distance of 850m. The car then braked and stopped in 28s. Calculate:

a) Time taken to reach 29ms-1 from 4ms-1

b) Its acceleration during this time

c) Distance it travelled during the deceleration

d) Its deceleration for the last 28s

by y.malik

a) Don’t know which suvat equation to use? Consider the variables given in each part of the question, and a method of elimination.

S=850

U=4.0

V=29

A=

T= ? We want to know the time taken from 4-29ms-1

v=u+at no s, therefore eliminate

s=ut+ ½ at2 no v, therefore eliminate

v2= u2+2as no v, therefore eliminate

s= ½ (u+v)t this has what we need s= ½ (u+v)t

850m = ½ (4+29)t

850 = 16.5t

t= 850/16.5 = 52s

by y.malik

b) Its acceleration during this time S=850 v2 = u2 + 2as U=4.0 292 = 42 +(2xax850) V=29 841=16+1700xa A= ? a = 841/1716 T= 52s a = 0.49ms-2

c) Distance it travelled during the deceleration

S= ? S = ½ (u+v)t U=29 = ½ (0+29) 28 V=0 s = 406m A= T= 28

d) Its deceleration for the last 28s S=406 v=u+at U=29 0 = 29+(ax28)

V=0 a= 29-28 A= ? a=-1ms-2 T= 28s

by y.malik

Q: cycle accelerates from rest at a constant acceleration of 0.4ms-2 for 20s then stops pedallying and slowed to standstill at constant deceleration over distance of 260m calculate:

a) Distance travelled by cyclist in the first 20s

b) The speed of the cyclist at end of this time

Answer

a) s= ut+ ½ at2 b) s= ½ (u+v)t

S= ? = ½ x0.4x202 S= 80m v= 8ms-1

U=0 s = 80m U= 0 OR

V=0.4 V = ? V=u+at

A= A = 0.4 v= 8ms-1

T=20 T=20

by y.malik

Q: a railwagon moving at speed of 2.0ms-1 on level track reached a steady incline which slowed it down to rest in 15s and caused it to reverse. Calculate:

a) Distance it moved up the incline

b) Its acceleration on the incline

Answer:

a) s= ½ (u+v)t b) s= ½ at2

s = 15m a= 0.13s

by y.malik