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%

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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%

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Answer 1 · 2017-07-31T17:59:32

$44 %$ $44%$

Explanation:

The applicable kinematic expression is

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

Since the car needs to stop, hence in the first instance distance traveled $s_{1}$ $s_1$ , when $r$ $r$ is the retardation

$0^{2} - u_{1}^{2} = - 2 r s_{1}$ $0^2-u_1^2=-2rs_1$
$\Rightarrow s_{1} = \frac{u_{1}^{2}}{2 r}$ $=>s_1=u_1^2/(2r)$

in the second instance given

$u_{2} = 1.2 u_{1}$ $u_2=1.2u_1$

Hence, distance traveled with same deceleration

$\Rightarrow s_{2} = \frac{{(1.2 u_{1})}_{1}^{2}}{2 r}$ $=>s_2=(1.2u_1)^2/(2r)$

Fractional increase in stopping distance

$\frac{s_{2} - s_{1}}{s_{1}} = \frac{\frac{{(1.2 u_{1})}_{1}^{2}}{2 r} - \frac{u_{1}^{2}}{2 r}}{\frac{u_{1}^{2}}{2 r}}$ $(s_2-s_1)/s_1=[(1.2u_1)^2/(2r)-u_1^2/(2r)]/(u_1^2/(2r))$
$\Rightarrow \frac{s_{2} - s_{1}}{s_{1}} = 1.44 - 1$ $=>(s_2-s_1)/s_1=1.44-1$
$\Rightarrow \frac{s_{2} - s_{1}}{s_{1}} = 0.44$ $=>(s_2-s_1)/s_1=0.44$

Percent increase in stopping distance $= 0.44 \times 100 = 44$ $=0.44xx100=44$

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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%

please give full explanation

1 Answer

Explanation:

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