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, the maximum velocity acquired by the car is given by
$(a)\left({\displaystyle \frac{{\alpha }^2+{\beta }^2}{\alpha \beta }}\right)t$
$\begin{array}{rl}& (b)\left({\displaystyle \frac{{\alpha }^2-{\beta }^2}{\alpha \beta }}\right)t\\ & (c)\left({\displaystyle \frac{\alpha +\beta }{\alpha \beta }}\right)t\\ & (d)\left({\displaystyle \frac{\alpha \beta }{\alpha +\beta }}\right)t\end{array}$

Answer 1

Hint: Assuming the maximum velocity, the one relation that relates all the given quantity is $v=u+at$. Using this and dividing the entire time span in two parts, the question can be solved easily.

Complete step by step answer:
Let us consider the car accelerates from rest at a constant rate $\alpha$ for time $t_{\alpha }$ to reach a maximum velocity of $v_{max}$. Therefore, we may write using the equation,

$v=u+at$

where, v is the final velocity, u is the initial velocity, a is the constant acceleration and t is the time taken. In this case,

$v=v_{max},u=0,a=\alpha ,t=t_{\alpha }$

Therefore, we get,

$v_{max}=0+\alpha \times t_{\alpha }$ …(I)

Again, let us consider, the car decelerates from $v_{max}$ to rest at a constant rate of $\beta$ for time $t_{\beta }$. Hence, we have,

$v=0,u=v_{max},a=-\beta ,t=t_{\beta }$

Therefore, we get,

$0=v_{max}-\beta \times t_{\beta }$ …(II)

Drawing the car’s journey on a graph, we get

From the above graph, AB is the car's acceleration journey and BC is the decelerated journey.

From the question we know, the entire time taken is t, where $t=t_{\alpha }+t_{\beta }$. Solving Eq. (I) and (II) for $t_{\alpha }$ and $t_{\beta }$, we get,

$t_{\alpha }={\displaystyle \frac{v_{max}}{\alpha }}$, $t_{\beta }={\displaystyle \frac{v_{max}}{\beta }}$

Therefore, substituting this value in the below equation, we get

$t=t_{\alpha }+t_{\beta }$
$t={\displaystyle \frac{v_{max}}{\alpha }}+{\displaystyle \frac{v_{max}}{\beta }}$

Now taking out the common term, we get

$t=v_{max}({\displaystyle \frac{1}{\alpha }}+{\displaystyle \frac{1}{\beta }})$

Taking LCM and solving, we get

$t=v_{max}\left({\displaystyle \frac{\alpha +\beta }{\alpha \beta }}\right)$
$v_{max}=\left({\displaystyle \frac{\alpha \beta }{\alpha +\beta }}\right)t$

Therefore, the correct option is D.

Note: If you look closely into the options, you would notice that they are of different dimensions. Hence, without even doing the sum, by performing a simple dimensional analysis, one can say the correct answer is D. It’s a trick that comes handy in a lot of such sums.