A body, moving at 20ms(^{-1}) accelerates uniformly at 2(frac{1}{2}ms^{-2}) for 4 seconds. It continues the journey at this speed for 8 seconds, before coming to rest seconds at tseconds after with uniform retardation. If the ratio of the acceleration to retardation is 3 : 4
(a( sketch the velocity – times graph of the journey
(b) find t
(c) find the total distance of the journey
Explanation
(a)
(b) If we let r be the retardation, then (frac{s}{t}) : r = 3 : 4
when simplified, r = (frac{10}{3})
Velocity after 4 seconds = 20 + (frac{5}{2}) x 4 = 30ms(^{-1})
So that (frac{5}{2}) : (frac{30}{t}) = (frac{3}{4})
t = 9 seconds
(c) The total distance of the journey = {[(frac{1}{2}) (20 + 30)4] + (30 x 8) + (frac{1}{2})(30 x 9)}
= 100 + 240 + 135 = 475 metres