QJEEM05Ph14ca: JEE Main 2005 Physics - Class 11 JEEM05Ph14ca Solution: A car starting from rest accelerates at the rate f through a…

Question

Verified Solution • 2025–26 Session moderate

A car starting from rest accelerates at the rate f through a distance S, then continues at constant speed for time t and then decelerates at the rate

{f \over 2}

to come to rest. If the total distance traversed is 15 S, then

Accepted Answer

Initially car starts from rest so u = 0.

Now distance from

A

to

B

,

\,\,\,\,\,\,\,\,\,\,

S = {1 \over 2}ft_1^2

\Rightarrow ft_1^2 = 2S

Distance from

B

to

C

= \left( {f{t_1}} \right)t

In B to C velocity is constant and v =

{f{t_1}}

Distance from

C

to

D

\,\,\,\,\,\,\,\,\,\,

= {{{u^2}} \over {2a}} = {{{{\left( {f{t_1}} \right)}^2}} \over {2\left( {f/2} \right)}} = ft_1^2 = 2S

\Rightarrow S + f\,{t_1}t + 2S = 15S

\Rightarrow f\,{t_1}t = 12S

........(1)

But

\,\,\,\,\,\,\,\,\,

{1 \over 2}f\,t_1^2 = S

.........(2)

On dividing the above two equations, we get

{t_1} = {t \over 6}

\Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}

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