Question

A particle moves on the x-axis and its velocity varies with time as $v\left(t\right)=t-t^2$. Find the time instant when the particle come back to starting point. Also find the distance travelled till then.

• A. t = 0 sec d = 1m
• B. t = 2 sec d = 2m
• C.
• D.

Answer 1

The correct option is C

Well we know v = $\frac{dx}{dt}$ where x-displacement

Now $\frac{dx}{dt}=t-t^2\Rightarrow \underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{x}{\int }}(t-t^2)dt$

Well the question asks us to find the time when the displacement of the particle becomes 0 i.e. at t the value of x should be 0

$\Rightarrow$ $\underset{0}{\overset{x}{\int }}dx$ $=\underset{0}{\overset{t}{\int }}(t-t^2)dt$

$x{]}_0^0=\frac{t^2}{2}-\frac{t^3}{3}{]}_t^0$

$0=\frac{t^2}{2}-\frac{t^3}{3}$

$\Rightarrow \frac{t^2}{2}=\frac{t^3}{3}$

$\Rightarrow \frac{1}{2}=\frac{t}{3}$

$\Rightarrow t=\frac{3}{2}sec$

$\Rightarrow$ at t = 1.5 sec particle comes back to mean position.

Now we have to find the distance travelled in the journey.

$v=t-t^2$

at t =1

Velocity of particle is 0

Till t = 1 sec the velocity is positive. After that velocity becomes negative that is the particle starts coming back towards origin and reaches back to starting point in next 0.5 sec, To find distance we need speed time relation.

We have velocity time graph. How will speed time look like

$\Rightarrow Speed=(t-t^2),fort=(0-1)sec$

$\Rightarrow =-(t-t^2),fort=(1-1.5)sec$

$\underset{0}{\overset{x}{\int }}dx=\underset{0}{\overset{1}{\int }}(t-t^2)dt-\underset{1.5}{\overset{1}{\int }}(t-t^2)dt$

$x=$ $[\frac{1}{2}-\frac{1}{3}]$$-[[\frac{9}{8}-\frac{9}{8}]-[\frac{1}{2}-\frac{1}{3}]]$

$=\frac{1}{3}m$