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GCSE Questions - Waves

Q15. The diagram shows Kayjay playing with a remote-controlled car.

(a) The graph below shows the distance-time variation for the first 30 seconds of the car's motion.

(i) Describe the motion of the car during the first 30 seconds.

It accelerates

[1 mark]

(ii) Determine the speed of the car 20 seconds after it started to move.

appropriate tangent drawn

correct reading from graph for change in distance and change in time (eg 5.6 (m) and 20 (s))

gradient of tangent worked out clearly

final answer of 0.28 m/s

[4 marks]

(b) A different car accelerated constantly from 0.12 m/s to 0.52 m/s.

The acceleration of the car was 0.040 m/s2 .

The work done to accelerate the car was 0.48 J.

Calculate the resultant force needed to accelerate the car.

As it is constant acceleration we can use Newton's equations of motion.

v2 - u2 = 2as

0.522 – 0.122 = 2 x 0.04 x s

s = 0.522 − 0.122 /(2 × 0.04)

s = 3.2 m

Work done = force x distance moved along the line of action of the force

W = Fs

0.48 = F x 3.2

F = 0.48/3.2

F = 0.15 N

Or you could use a different Newton equation:

v = u + at

t = (v - u)/a

t = (0.52 – 0.12)/0.04

t = 10 s

average speed = (v + u)/2

= (0.52 + 0.12)/2

= 0.64/2

= 0.32 m/s

distance travelled = average speed x time

s = 0.32 x 10 = 3.2 m

W = Fs

0.48 = F x 3.2

F = 0.48/3.2

F = 0.15 N

Or you could use the change in kinetic energy to calculate it:

Δ Energy = ½mv2 - ½mu2

0.48 = (0.5 x m x 0.522) – (0.5 x m x 0.122)

0.48 = 0.1352m – 0.0072m

0.48 = 0.128m

m = 3.75 kg

F = ma

F = 3.75 x 0.040

F = 0.15 N

 

[6 marks]

(c) Explain why the car has a maximum speed.

There is a maximum possible forward force provided by the motor. As the speed of the car increases air resistance increases until air resistance is equal in size to that forward force, then the car can no longer accelerate as it has reached its terminal velocity.

[4 marks]

(Total 15 marks)