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...

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

Answer Details

a)

${}^{b)}$

If we let r be the retardation, then st

$\frac{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">𝑠</span>}}{\mathrm{<span\; style="font-family:\; Ubuntu;\; font-size:\; 16px;">𝑡</span>}}$ : r = 3 : 4

when simplified,

r = 103 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

Velocity after 4 seconds

= 20 + 52 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ x 4 = 30ms−1 ${}^{-1}$

So that 52 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ : 30t $\frac{\mathrm{<br>}}{\mathrm{<br>}}$ = 34 $\frac{\mathrm{<br>}}{\mathrm{<br>}}$

t = 9 seconds