Question
An automobile travelling with speed of 60 km/h, can brake to stop within a distance of 20 cm. If the car is going twice as fast, i.e 120 km/h, the stopping distance will be
-
20 m
-
40 m
-
60 m
-
80 m
Solution
D.
80 m
Third equation of motion gives
v2 = u2 + 2as ⇒ 2
s ∝ u (since v = 0)
where a = retardation of body in both the cases
therefore, 
.... (i)
Here, s1 = 20 m, u1 = 60 km/h, u2 = 120 km/h. Putting the given values in eq. (i), we get
