Question

A car accelerates from rest with a constant acceleration $\alpha$ on a straight road. After gaining a velocity $v$, the car moves with the same velocity for some time. Then the car is decelerated to rest with a retardation $\beta$. If the total distance covered by the car is equal to $S$, the total time taken for its motion is

• A. $\frac{S}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
• B. $\frac{S}{v}+\frac{v}{\alpha }+\frac{v}{\beta }$
• C. $(\frac{v}{\alpha }+\frac{v}{\beta })$
• D. $\frac{S}{v}-\frac{v}{2}(\frac{v}{\alpha }+\frac{v}{\beta })$

Answer 1

The correct option is A $\frac{S}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$

Let’s try to solve this problem with the help of graphical representation. Figure below illustrates the situation given in the problem

From $v=u+at$, we have
When the body is accelerating
$v=0+\alpha t_1\Rightarrow t_1=v/\alpha .......\left(i\right)$
When the body is decelerating
$0=v-\beta t_3\Rightarrow t_3=v/\beta ......\left(ii\right)$
From the figure and from $\left(i\right)$ and $\left(ii\right)$, we can say
$S=\frac{1}{2}v{t}_1+v{t}_2+\frac{1}{2}v{t}_3$
$=\frac{v^2}{2\alpha }+v{t}_2+\frac{v^2}{2\beta }$
$\Rightarrow t_2=\frac{S}{v}-\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
Using $\left(i\right),\left(ii\right)$ and $\left(iii\right)$, we get
$t_1+t_2+t_3=\frac{S}{v}+\frac{v}{2}(\frac{1}{\alpha }+\frac{1}{\beta })$
Thus, the correct answer is option (a)