Question:

A car moving along a straight highway with speed of 126 Kmh^-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?

Solution:

Initial velocity of the car u =126 Kmh^-1

Converting the velocity into meter per second

$${126\frac{Km}{Hr}}={126\frac {(1000 \ m)}{(3600 \ s)}} ={126 \frac{(5)}{(18)}} ={35\frac{m}{s}}$$

Final velocity of the car v=0 (Brought to a stop)

Distance covered s= 200 m

Retardation of the car:

Applying third equation of motion

$${v^2 =u^2+2as} $$

$${a =\frac{v^2-u^2}{2s}}$$

$${a=\frac{(0^2)-(35^2)}{400}}$$

$${a= -\frac{1225}{400}}$$

$${a=-3.06 \ \frac{m}{s^2}}$$

How long does it take for the car to stop?

$${v=u+at}$$

$$t=\frac{v-u}{a}$$

$${t=\frac{0-35}{-3.06}}$$

$${t=11.44 \ s}$$