Answer:
The required skidded distance would be one-forth of the distance skidded if the velocity of the car becomes half.
Explanation:
If '[tex]v_{0}[/tex]' 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 '[tex]v_{t} = 0[/tex]', then as can be seen from the figure,
[tex]&& v_{t} = v_{0} - a~t\\&or,& 0 = v_{0} - a~t\\&or,& t = \dfrac{v_{0}}{a}[/tex]
If from any arbitrarily chosen point 'O', '[tex]x_{0}[/tex]' be the distance where the brakes are slammed on and '[tex]x_{t}[/tex]' be the distance from the same point 'O' where the car stops, then
[tex]&& 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}[/tex]
Now if the car would have been moving with a velocity '[tex]\dfrac{v_{0}}{2}[/tex]' and would have been skidding a distance 'Y', then from the above equation substituting the velocity we can have
[tex]Y = \dfrac{X}{4}[/tex]
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.
