Question

A car moves at a speed of 50 kilometers/hour. Its kinetic energy is 400 joules. If the same car moves at a speed of 100 kilometers/hour, then its kinetic energy will be joules. The braking distance at the faster speed is the braking distance at the slower speed.

Answer 1

Explanation :

Speed of car, $v_1=50\ km/h=13.8\ m/s$

Kinetic energy of the car, $KE_1=400\ J$

Speed of car, $v_2=100\ km/h=27.7\ m/s$

Let $KE_2$ is the kinetic energy of the car when it is moving with 100 km/h.

$\dfrac{KE_1}{KE_2}=\dfrac{\dfrac{1}{2}mv_1^2}{\dfrac{1}{2}mv_2^2}$

$\dfrac{KE_1}{KE_2}=\dfrac{v_1^2}{v_2^2}$

$KE_2=KE_1(\dfrac{v_2}{v_1})^2$

$KE_2=400\ J\times (\dfrac{27.7\ m/s}{13.8\ m/s})^2$

$KE_2=1611.6\ J$  

Initial velocity of both cars are 0. Using third equation of motion :

So, $S_1=\dfrac{v_1^2}{2a}=\dfrac{v_1t}{2}=\dfrac{v_1t}{2}$

and $S_2=\dfrac{v_2^2}{2a}=\dfrac{v_2t}{2}=\dfrac{v_2t}{2}$

t is same. So,

$\dfrac{S_1}{S_2}=\dfrac{50\ m/s}{100\ m/s}$

$\dfrac{S_1}{S_2}=\dfrac{1}{2}$

$S_2=2\ S_1$

So, the braking distance at the faster speed is twice the braking distance at the slower speed.

Answer 2

I just took this test its 1,600 and quadruple.