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# A car of mass m moving at a speed v is stopped in distance x by friction between tyres and road. If kinetic energy of the car is doubled. its stopping distance will be

$(a)\;8\;x \quad (b)\;4\;x \quad (c)\;x \quad (d)\;2\;x$

If a is retardation due to friction f
$2ax=v^2$
Kinetic energy $=\large\frac{1}{2} $$mv^2 \qquad=\large\frac{1}{2}$$ m (2ax)=max$
$K.E \;\alpha\; x$
Thus if kinetic energy is double distance is also doubled.
Hence  c is the correct answer.

edited Feb 10, 2014 by meena.p