Question

A car moving along a straight highway with speed of 126 km h ^-1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform) and how long does it take for the car to stop?

Answer 1

Initial velocity of car, u = 126 kmh^-1 = 126 x \(\frac{5}{18}\) ms^-1 = 35 ms^-1 ………(i) 

Since, the car finally comes to rest, v = 0 

Distance covered, s = 200 m, a = ?, t = ?

v² = u² – 2as

or a = \(\frac{v^2 - u^2}{2s}\) ………..(ii)

substituting the values from eq. (i) in eq . (ii), we get

Negative sign shows that acceleration in negative which is called retardation, i.e., car is uniformly retarded at – a = 3.06 ms^-2

To find t, let us use the relation 

v = u + at

t = \(\frac{v-u}{a}\)

use a = -3.06 ms^-2 , v = 0, u = 35 ms^-1

∴ t = \(\frac{v-u}{a}\) = \(\frac{0-35}{-3.06}\) = 11.44 s

∴ t = 11.44 sec