If the velocity of car is increased by 20% then minimum distance in which it can be stopped increases by ? 1. 44% 2. 55% 3. 66% 4. 88%

Question

please give full explanation

10479 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-31T17:59:32

$44%$

The applicable kinematic expression is

$v^2-u^2=2as$
where $v$ is final velocity, $u$ is initial velocity, $a$ is acceleration and $s$ is displacement.

Since the car needs to stop, hence in the first instance distance traveled $s_1$ , when $r$ is the retardation

$0^2-u_1^2=-2rs_1$
$=>s_1=u_1^2/(2r)$

in the second instance given

$u_2=1.2u_1$

Hence, distance traveled with same deceleration

$=>s_2=(1.2u_1)^2/(2r)$

Fractional increase in stopping distance

$(s_2-s_1)/s_1=[(1.2u_1)^2/(2r)-u_1^2/(2r)]/(u_1^2/(2r))$
$=>(s_2-s_1)/s_1=1.44-1$
$=>(s_2-s_1)/s_1=0.44$

Percent increase in stopping distance $=0.44xx100=44$