Question

A car moving with a velocity of 20 m/s stopped in a distance of 40 m. If the same car is travelling at double the velocity, the distance travelled by it for same retardation is
(1) 320 m (2) 1280 m (3) 160 m (4) 640 m
​

Answer 1

Given : initial velocity of car = 20 m/s

            distance travelled by car before stopping = 40 m

To Find : distance travelled by car when it has double the initial velocity

Solution:

• For solving this question we can use the formula:

                            $v^{2}$ - $u^{2}$ = 2as                                     (1)

       where, v is the final velocity of car = 0 m/s       (it stops)

                    u is the initial velocity if car = 20 m/s

                    a is the retardation                               ( to be calculated)

             and s is the distance travelled = 40 m

Substituting all these values in equation (1),

                        ⇒     $0^{2}$ - $20^{2}$ = 2a × 40

                        ⇒      -400 = 80 × a

                       ⇒          a = $\frac{-400}{80\\}$

                        ⇒         a = -5 m/$s^{2}$              (2)

Here, acceleration has negative sign as it is acting in the direction opposite to motion of car (retardation).

• For second case,

                               initial velocity u' = 2u = 2  × 20 = 40 m/s                  

                               final velocity v' = 0 m/s

                                retardation a = -5 m/$s^{2}$         (same as (2) as per question)

                                distance travelled = s'           (to be calculated)

                   Substituting these values in equation (1)

                             ⇒  $0^{2}$ - $40^{2}$ = 2 × (-5) × s'

                             ⇒  -1600 = -10 × s'

                             ⇒   s' = $\frac{-1600}{-10}$

                             ⇒ s' = 160 m

∴ For the same retardation and double velocity, the car will travel 160 m hence option (3) 160 m is correct.