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Q1
The graph below shows the distance-time graph for the first 30 seconds of the car’s motion.
(f) A different car accelerated 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.
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Resultant force = _______________ N (6)
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Final velocity² – Initial velocity² = 2 × acceleration × distance
0.522−0.122=2×0.04×s
s=0.52^2−0.12^2/2×0.04
s=0.2704−0.0144/0.08
s=0.256/0.08=3.2 m
Work done = Force × Distance
0.48=F×3.2
F=0.48/3.2=0.15 N
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