Question

A car accelerates from rest at a constant rate $alpha$ for some time after which it decelerates at a constant rate $\beta$ to come to rest. If the total time elapsed is t sec, evaluate the total distance travelled.

Answer 1

When initial velocity is 0 m/s let time taken to accelerate be $t_1$ and final velocity be v, thus acceleration $\alpha$ can be

written as $\alpha =\frac{v}{t_1}$

$t_1=\frac{v}{\alpha }$------(1)

Similarly for acceleration $\beta$ at time $t_2$ final velocity is 0 m/s and initial velocity will be equal to the final velocity of previous condition viz. $u=v$

Thus $\beta =\frac{-v}{t_2}$

$t_2=\frac{-v}{\beta }$-----(2)

Now $t_1+t_2=t$

So adding 1 and 2 we get

$t_1+t_2=\frac{v}{\alpha }-\frac{v}{\beta }$

$t=\frac{\beta v-\alpha v}{\alpha \beta }$

Also we know $velocityv=\frac{s}{t}$

$s=\frac{\beta v^2-\alpha v^2}{\alpha \beta }$