Question 91: Motion in 1D

Question

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 $${t_1} = {t \over 6}$$
$$ \Rightarrow S = {1 \over 2}f{\left( {{t \over 6}} \right)^2} = {{f\,{t^2}} \over {72}}$$

Answer

$$S = {1 \over 6}f{t^2}$$

Answer

$$S = ft$$

Answer

$$S = {1 \over 4}f{t^2}$$

Answer

$$S = {1 \over 72}f{t^2}$$

Motion in 1D

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

Choose the correct answer: