Question

A car accelerates from rest at a constant rate of 2 m/s2 for some time, after that it retards at a constant rate of 3 m/s2 and come to rest. If the total time elapsed is 8 s, then the maximum velocity acquired by the car is


9.6 m/s


4.8 m/s


2.4 m/s


12 m/s

Answer 1

Answer:

The maximum velocity acquired by the car is 9.6 m/s.

Explanation:

Given,

$a_{1} = 2 m/s^{2}$

$a_{2} = 3 m/s^{2}$

The total time elapsed (t) = 8 s

Let,

v = The maximum velocity attained.

$t_{1}$ = Time for which car accelerated

$t_{2}$ = Time for which car retarded.

∴$t_{1} = \frac{v-u}{a_{1} }$                           and    $t_{2} = \frac{v-u}{a_{2} }$

      $= \frac{v-0}{2}$                                          $= \frac{0-v}{-3}$

      $=\frac{v}{2}$                                              $=\frac{v}{3}$

Now,

The total time elapsed (t) = 8 s

⇒ $t_{1} + t_{2} = 8$

⇒ $\frac{v}{2} +\frac{v}{3} =8$

⇒ $\frac{3v+2v}{6} = 8$

⇒ 5v = 48

 ∴ v = 9.6 m/s

The maximum velocity acquired by the car is 9.6 m/s.