Question

If the speed of a car is increased by 80%, by what factor will its minimum braking distance be increased, assuming all else is the same? Ignore the driver's reaction time.

Answer 1

Answer:

$KE_2=3.24 \times KE_1$

Explanation:

given,

Speed of car = v

new speed of car v'= 1.8 v

Kinetic energy of the car

$KE_1 =\dfrac{1}{2}mv^2$

new Kinetic energy of energy

$KE_2 =\dfrac{1}{2}mv'^2$

$KE_2=\dfrac{1}{2}m(1.8v)^2$

$KE_2=3.24 \times \dfrac{1}{2}mv^2$

$KE_2=3.24 \times KE_1$

hence, the kinetic energy is increased by the factor of 3.24.