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%#

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#