Mar 5, 2016

$4 \frac{m}{s}$

Explanation:

The car starts from rest therefore its initial velocity is zero, i.e., ${v}_{i} = 0$ in case when its acceleration is ${a}_{1} = 2 \frac{m}{s} ^ 2$.

Let the car come to a final velocity ${v}_{f} = v$. in time ${t}_{1}$

Then we can write:

${v}_{f} = {v}_{i} + {a}_{1} {t}_{1}$

$\implies v = 0 + 2 {t}_{1}$

$\implies v = 2 {t}_{1}$

$\implies {t}_{1} = \frac{v}{2.} \ldots \ldots \ldots \ldots \ldots . \left(i\right)$

Now when it is again coming to rest its initial velocity is that which it attained when it started from rest i.e., $v$

Hence, when it is again coming to rest at that period ${v}_{i} = v$, ${v}_{f} = 0$ and ${a}_{2} = - 4 \frac{m}{s} ^ 2$ (NOTE: The negative sign for acceleration is taken because it is retardation). Let the time which it took for coming to rest from the velocity $v$ be ${t}_{2}$.

Thus, we can write:

${v}_{f} = {v}_{i} + {a}_{2} {t}_{2}$

$\implies 0 = v - 4 {t}_{2}$

$\implies v = 4 {t}_{2}$

$\implies {t}_{2} = \frac{v}{4.} \ldots \ldots \ldots \ldots . . \left(i i\right)$

Adding equations $\left(i\right)$ and $\left(i i\right)$, we get.

${t}_{1} + {t}_{2} = \frac{v}{2} + \frac{v}{4}$

${t}_{1} + {t}_{2}$ represents the total time for this trip i.e., starting from rest and then again coming to rest.
And it is given that the total time of trip is $3$ seconds.

$\implies 3 = \frac{v}{2} + \frac{v}{4}$

$\implies 12 = 2 v + v$

$\implies 3 v = 12$

$\implies v = 4 \frac{m}{s}$

Hence, the maximum velocity the car attained is $4 \frac{m}{s}$.

Both the given options in the question are wrong.