Question

A body starts from rest with acceleration $2{\text{m/s}}^2$ till it attains the maximum velocity, then retards to rest with $3{\text{m/s}}^2$. If total time taken is $10$ seconds then maximum speed attained is

• A. $12\text{m/s}$
• B. $8\text{m/s}$
• C. $6\text{m/s}$
• D. $4\text{m/s}$

Answer 1

The correct option is A $12\text{m/s}$

$\text{I}$ - Method
Let the maximum velocity attained at time $\left(t^{\prime }\right)$ be $v_{max}$.
By first equation of motion

$v_{max}=u+a{t}^{\prime }\Rightarrow v_{max}=2t^{\prime }...\left(\text{i}\right)$

Now for the second part of the journey body comes to rest $(v=0)$ and the time will be $(10-t^{\prime })$ as the total journey time is $10\text{s}$, $u=v_{max}$, $a=-3{\text{m/s}}^2$. By first equation of motion

$0=v_{max}-3(10-t^{\prime })...\left(\text{ii}\right)$ putting the value of $v_{max}$ in equation $\left(\text{ii}\right)$ we get,
$2t^{\prime }=3(10-t^{\prime }),\Rightarrow t^{\prime }=6\text{s}$
$v_{max}=2t^{\prime }=2\times 6=12\text{m/s}$

$\text{II}-$Method
As we know that slope $\left(\tan \theta \right)$ of velocity-time $(v-t)$ graph gives acceleration.
$\tan \alpha =2,\tan \beta =3$
$2=\frac{v}{t},3=\frac{v}{10-t}$
$v=2t,v=30-3t$
$30-3t=2t\Rightarrow t=6\text{sec}$
$v=12\text{m/s}$