Question

A particle starts from rest and travels a distance $l$ with uniform acceleration, then moves uniformly over a further distance $2l$ and finally comes to rest after moving a further distance $3l$ under uniform retardation. Assuming entire motion to be rectilinear motion the ratio of average speed over the journey to the maximum speed on its ways is

• A. $1/5$
• B. $2/5$
• C. $3/5$
• D. $4/5$

Answer 1

The correct option is C $3/5$

Let $v_m$ be the maximum speed of the particle
Let $t_1,t_2$ and $t_3$ be the time taken by the particle for first, second and third motion respectively.
Let $a_1$ and $a_2$ are the accelerations of first and second motion respectively and $a_3$ is the retardation of third motion(because particle is coming to rest)
When the particle covers distance $l$
By using first equation of motion, $v_m=u+a_1{t}_1$
$v_m=a_1{t}_1$
$\Rightarrow t_1=\frac{v_m}{a_1}$
Using third equation of motion, $v_m=\sqrt{2a_{1}l}$
$a_1=\frac{v_m^2}{l}$
When the particle covers $2l$ distance with constant speed i.e $(a_2=0)$
$\Rightarrow t_2=\frac{2l}{v_m}$
When the particle covers distance $3l$
Using first equation of motion, $0=v_m-a_3{t}_3$
$v_m=a_3{t}_3$
$t_3=\frac{v_m}{a_3}$
Using third equation of motion, $0^2=v_m^2-2a_3\left(3l\right)$
$v_m=\sqrt{2a_3\left(3l\right)}\Rightarrow a_3=\frac{v_m^2}{6l}$
Now, average speed: $v_{avg}=\frac{l+2l+3l}{t_1+t_2+t_3}$
Now substitute $t_1,t_2$ and $t_3$ in above equation we get
$v_{avg}=\frac{6l}{\left(v_m/a_1\right)+\left(2l/v_m\right)+\left(v_m/a_3\right)}$
$=\frac{6l}{\left(\frac{v_m}{v_m^2/2l}\right)+\left(\frac{2l}{v_m}\right)+\left(\frac{v_m}{v_m^2/6l}\right)}\Rightarrow \frac{6l{v}_m}{2l+2l+6l}=\frac{6l{v}_m}{10l}$ $\frac{v_{avg}}{v_m}=\frac{3}{5}$