Question

A car starts from rest and travels with uniform acceleration $\alpha$ for some time and then with uniform retardation $\beta$ and comes to rest. If the total time of travel of the car is t, then the maximum velocity attained by the car is given by
$A){\text{ }}\left( {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right)t \\\ B){\text{ }}\dfrac{1}{2}\left( {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right){t^2} \\\ C)\left( {\dfrac{{\alpha \beta }}{{\alpha - \beta }}} \right)t \\\ D)\dfrac{1}{2}\left( {\dfrac{{\alpha \beta }}{{\alpha - \beta }}} \right){t^2} \\\$

Answer 1

we have to calculate the velocity with the help of Newton’s equation of motion
Here’s the expression we used
$v = u + at$

Complete step by step solution:
A car starts from rest means initial velocity is zero i.e., $u=0m/s$
Acceleration, a= $\alpha m/{s^2}$
Retardation, a= $\beta m/{s^2}$ (retardation is nothing but is a negative acceleration or deceleration)
Now, we calculate the velocity,
From the expression,
$v = u + at$
We find the velocity for different acceleration with different interval of time
For uniform acceleration time be
Expression be
$v = u + at$
Now, put the value of initial velocity (u), acceleration (a) and time (t) in above expression
$v = 0 + \alpha {t_1}$
$\Rightarrow v = \alpha {t_1}$ -- (A)
For uniform retardation time be
Expression be
$v = u + at$
Now, put the value of initial velocity (u), acceleration (a) and time (t) in above expression
$v = 0 + \beta {t_2}$
$\Rightarrow v = \beta {t_2}$ --- (B)
We find the both velocity now we calculate the total time taken,
Here’s the expression
Total time taken = time taken for uniform acceleration + time taken for uniform retardation
$t = {t_1} + {t_2}$ -- (C)
From eq. (A) we have to calculate the time interval for uniform acceleration
$v = \alpha {t_1} \\\ \Rightarrow {t_1} = \dfrac{v}{\alpha } \\\$
From eq. (B) we have to calculate the time interval for uniform retardation.
$v = \beta {t_2} \\\ \Rightarrow {t_2} = \dfrac{v}{\beta } \\\$
Now, put the value of ${t_1}$ and ${t_2}$ in eq. (C)
$t = {t_1} + {t_2} \\\ \Rightarrow t = \dfrac{v}{\alpha } + \dfrac{v}{\beta } \\\ \Rightarrow t = v\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right) \\\ \\\$
Here, taking common v , because we have to find the maximum velocity attained
$t = v\left( {\dfrac{1}{\alpha } + \dfrac{1}{\beta }} \right) \\\ \Rightarrow t = v\left( {\dfrac{{\beta + \alpha }}{{\alpha \beta }}} \right) \\\ \Rightarrow v = \left( {\dfrac{{\alpha \beta }}{{\alpha + \beta }}} \right)t \\\$
Hence, the option (A) is correct.

Note: We have to find the time with the help of Newton’s law of motion. We can also find the value of time by the definition of acceleration. We know that expression:

$$acceleration = \dfrac{{velocity}}{{time}} \\\ So, {\text{ time = }}\dfrac{{velocity}}{{acceleration}} \\\$$