A body, moving at 20ms\(^{-1}\) accelerates uniformly at 2\(\frac{1}{2}ms^{-2}\) for 4 seconds. It continues the...
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
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(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