Answer:
option D
Explanation:
given,
mass of the car 1 = m
speed of car 1 = v
time taken to stop the car = t
mass of car 2 = m' = 2 m
speed of the car 2 = v
time taken by the car to stop = t' = ?
now,
we know,
F = m a....(1)
and force by the second car
F = m' a'
F = 2 m a'
[tex]a' = \dfrac{F}{2m}[/tex]
from equation (1)
[tex]a' = \dfrac{a}{2}[/tex]
using equation of motion
v = u + at
0 = v - a t
[tex]t = \dfrac{v}{a}[/tex]
again using equation of motion for the calculation of the time taken by the second car.
[tex]t'= \dfrac{v}{a'}[/tex]
[tex]t' = 2\dfrac{v}{a}[/tex]
t' = 2 t
hence, the time taken by the second car is twice the time taken by the first car to stop.
The correct answer is option D