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?
$${v=u+at}$$
$$t=\frac{v-u}{a}$$
$${t=\frac{0-35}{-3.06}}$$
$${t=11.44 \ s}$$
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