A mechanic checks the stopping force of a small car on a test track. Fig. 1 shows the car travelling at 18 m/s before braking. Its mass is 950 kg. During a ...
Question 1 Report
A mechanic checks the stopping force of a small car on a test track. Fig. 1 shows the car travelling at 18 m/s before braking. Its mass is 950 kg. During a braking time of 3.0 s, its velocity decreases uniformly to 6.0 m/s.
(a) Calculate the acceleration of the car while braking. [2]
(b) Calculate the resultant force on the car. [2]
(c) State the direction of this resultant force. [1]
(d) Calculate the distance moved by the car during the 3.0 s braking time. [3]
(e) Explain why a passenger moves forward relative to the car when it starts to brake. [2]
Answer Details
(a) Acceleration is change in velocity divided by time:
\[a=\frac{v-u}{t}=\frac{6.0-18}{3.0}=-4.0\text{ m/s}^2\]
The acceleration is \(-4.0\text{ m/s}^2\). The negative sign shows that the acceleration is opposite to the original direction of motion. [2]
(b) Apply Newton's second law, \(F=ma\):
\[F=950\times(-4.0)=-3800\text{ N}\]
The resultant force is \(-3800\text{ N}\), with magnitude 3800 N. [2]
(c) The resultant force acts opposite to the direction of motion, so it acts backwards. [1]
(d) With uniform slowing, mean speed is:
\[\text{mean speed}=\frac{18+6.0}{2}=12\text{ m/s}\]
\[s=\text{mean speed}\times t=12\times3.0=36\text{ m}\]
The car moves 36 m. [3]
(e) The passenger has inertia, meaning that they tend to continue at their original velocity. [1] When the car slows, the passenger therefore continues forwards relative to the car until a force, such as from the seat belt, slows them. [1]
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