Question

If the velocity of car is increased by 20% then the minimum distance in which it can be stopped increases by

Answer 1

Answer: 44%

Solution:

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

$0^{2} - u_{1} ^{2} =-2r s_{1}$
⇒$s_{1} = \frac{u_{1} ^{2}}{2r}$

in the second instance given

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

Hence, distance traveled with same deceleration

⇒$s_{2}= \frac{ (1.2 u_{1} )^{2} }{2r}$

Fractional increase in stopping distance

$\frac{ s_{2} -s_{1}}{s_{1}}= \frac{\frac{ (1.2 u_{1} )^{2} }{2r}-\frac{ u_{1} ^{2} }{2r}}{ \frac{ u_{1} ^{2} }{2r} }$
⇒$\frac{ s_{2} -s_{1}}{s_{1}}=1.44-1=0.44$

∴Percent increase in stopping distance=0.44×100%=44%

Answer 2

Answer:

Step-by-step explanation: i dont know the answer but al the answers are complocated