Question

You slam on the brakes of your car in a panic and skid a distance X on a straight, level road. If you had been traveling half as fast under the same road conditions, you would have skidded a distance

Answer 1

Answer:

The required skidded distance would be one-forth of the distance skidded if the velocity of the car becomes half.

Explanation:

If '$v_{0}$' be the velocity of the car where the brakes are slammed on and 'a' be the constant deceleration gained by the car before coming to stop after skidding a distance 'X' at time 't' where final velocity '$v_{t} = 0$', then as can be seen from the figure,

$&& v_{t} = v_{0} - a~t\\&or,& 0 = v_{0} - a~t\\&or,& t = \dfrac{v_{0}}{a}$

If from any arbitrarily chosen point 'O', '$x_{0}$' be the distance where the brakes are slammed on and '$x_{t}$' be the distance from the same point 'O' where the car stops, then

$&& x_{t} = \int\limits^t_0 {v_{t}} \, dt\\ = \int\limits^t_0 ({v_{0} - a~t}) \, dt\\ = v_{0}~t - \dfrac{1}{2}~a~t^{2} + x_{0}\\\\&or,& x_{t} - x_{0} = v_{0}~t - \dfrac{1}{2}~a~t^{2}\\&or,& X = v_{0}~\dfrac{v_{0}}{a} - \dfrac{a}{2}(\dfrac{v_{0}}{a})^{2}\\&or,& X = \dfrac{v_{0}^{2}}{a}$

Now if the car would have been moving with a velocity '$\dfrac{v_{0}}{2}$' and would have been skidding a distance 'Y', then from the above equation substituting the velocity we can have

$Y = \dfrac{X}{4}$

So, if the car were moving at half as fast under the same road conditions, it would have skidded by a distance which is one-forth of the present skidded distance.