(a) Define uniform acceleration.
(b) Forces act on a car in motion. List the
(i) horizontal forces and their directions;
(ii) vertical forces and their directions
(c) A car starts from rest and accelerate uniformly for 20s to attain a speed of 25 ms(^{-1}). It maintains this speed for 30s before decelerating uniformly to rest. The total time for the journey is 60s.
(i) Sketch a velocity-tune graph for the motion.
(ii) Use the graph to determine the ((alpha)) total distance travelled by the car ((beta)) deceleration of the car.
The figure here illustrates force-extension graph for a stretched spiral spring. Determine the work done on the spring.
Explanation
(a) Uniform acceleration is the constant time rate of increase in velocity.
(b) (i) Horizontal Force;
Forward force – Thrust
backward force – Frictional force
(ii) Vertical Force
Upward force – Normal reaction.
Downward force – Weight of the car.
(c)
(ii) ((lambda)) Total distance
= Area of trapezium OABC
= (frac{1}{2})[AB + OC] AD
= (frac{1}{2})[60+30] x 25
=1125m
((beta)) Deceleration = slope of BC
= (frac{BE}{CE} = frac{25}{10})
= 25m/s(^2)
(d) Work done = (frac{1}{2})Fe
= (frac{1}{2}) x 12 x (frac{0.5}{100})J
= 0.03J