Answer:
one forth.
Explanation:
car is moving at the speed of v
car stops, final speed = s
distance to stop the car = ?
using equation of motion
u_i² = u_f² + 2 a s
0² = v² - 2 a s
-ve sign is used because the car is decelerating.
[tex]s = \dfrac{v^2}{2a}[/tex]
now, if the velocity of the car is 2v distance to stop
[tex]s' = \dfrac{(2v)^2}{2a}[/tex]
[tex]s' = 4\dfrac{v^2}{2a}[/tex]
[tex]s' = 4 s[/tex]
[tex]s = \dfrac{s'}{4}[/tex]
now, the distance is one forth.
so, car with speed v has to cover one forth of the distance cover by car with speed 2 v.
