(a) Explain the terms:- uniform acceleration and average speed.
(b) A body at rest is given an initial uniform acceleration of 8.0ms(^{-2}) for 30s after which the acceleration is reduced to 5.0ms(^{-2}) for the next 20s. The body maintains the speed attained for 60s after which it is brought to rest in 20s. Draw the velocity-time graph of the motion using the information given above.
(c) Using the graph, calculate the: (i) maximum speed attained during the motion; (ii) average retardation as the body is being brought to rest; (iii) total distance travelled during the first 50s; (iv) average speed during the same interval as in (ii).
Explanation
(a) Uniform acceleration occurs when the change in velocity per unit time is constant. The S.I. unit is ms(^{-2}).
(a = frac{v – u}{t}); a – uniform acceleration, v – final velocity, u – initial velocity, t – time.
Graphically,
acceleration is the same acceleration at x is different every point from that at Y.
Note – acceleration is given by slope in both graphs.
(b)
(i) Maximum speed = 240 m/s.
(ii) (a = frac{v – u}{t} = frac{0 – 200}{20} = -10 ms^{-2})
(iii) Area of I = (frac{1}{2} times 30 times 240 = 3600m)
Area of II = (20 times 200 = 4000m)
Area of III = (frac{1}{2} times 20 times 40 = 400m)
Total distance in the first 50s : (3600m + 4000m + 400m = 8000m)
(iv) Average speed during the first 50s = (frac{8000m}{50s} = 160 ms^{-1})