(a) (i) Define force and state its S.I unit.
(ii) List the two types of solid friction.
(b) A car travelling at a constant speed of 30 ms-1 for 20 s was suddenly decelerated when the driver sighted a pot-hole. It took the driver 6 s to get to the pot-hole with a reduced speed of 18 ms-1. He maintained the steady speed for another 10 s to cross the pot-hole. The brakes were then applied and the car came to rest 5 s later.
(i) Draw the velocity-time graph for the journey.
(ii) Calculate the deceleration during the last 5 s of the journey.
(a)
(i) Force is a push or pull acting on an object that causes it to accelerate or deform. The SI unit of force is newton (N).
(ii) The two types of solid friction are static friction and kinetic friction.
(b)
(i)
The velocity-time graph for the journey is shown above. The horizontal axis represents time in seconds, and the vertical axis represents velocity in meters per second (m/s). The graph shows that the car was travelling at a constant velocity of 30 m/s for 20 seconds, then decelerated to a velocity of 18 m/s over a period of 6 seconds, maintained that velocity for 10 seconds, and finally came to rest over a period of 5 seconds.
(ii)
To calculate the deceleration during the last 5 seconds of the journey, we can use the formula:
a = (vf - vi) / t
where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval. In this case, the final velocity is 0 m/s, the initial velocity is 18 m/s, and the time interval is 5 seconds. Therefore:
a = (0 - 18) / 5 = -3.6 m/s2
The negative sign indicates that the acceleration is in the opposite direction to the initial velocity, which is consistent with deceleration.
(iii)
To calculate the total distance covered, we need to break the journey down into three parts:
- The distance covered while travelling at a constant velocity of 30 m/s for 20 seconds.
- The distance covered while decelerating from 30 m/s to 18 m/s over a period of 6 seconds, and then maintaining a constant velocity of 18 m/s for 10 seconds.
- The distance covered while coming to rest over a period of 5 seconds.
For part 1, the distance covered is:
distance = velocity x time
distance = 30 m/s x 20 s = 600 m
For part 2, we need to calculate the average velocity over the 16-second period (6 seconds of deceleration + 10 seconds of constant velocity). The average velocity is:
average velocity = (initial velocity + final velocity) / 2
average velocity = (30 m/s + 18 m/s) / 2 = 24 m/s
The distance covered during this 16-second period is:
distance = velocity x time
distance = 24 m/s x 16 s = 384 m
For part 3, the distance covered is:
distance = (final velocity / 2) x time
distance = (0 m/s / 2)
