Question

The velocity of a particle starting from origin varies with time as $v=2t+1$. Find the position, x at any time instant t?
(a) $x={{t}^{2}}+t$
(b) $x=2{{t}^{2}}+t$
(c) $x=4{{t}^{2}}+t$
(d) $x={{t}^{2}}+1$

Answer 1

Hint: Use the information that the velocity of a particle along a direction is the derivative of displacement with respect to time along that direction and replace $v$ with $\dfrac{dx}{dt}$ in the relation $v=2t+1$. Multiply both the sides with $dt$ and integrate both the sides with suitable limits to get the position x. Take lower limit in the L.H.S as 0 and upper limit as x. Similarly, take the lower limit in the R.H.S as 0 and upper limit as t.

Complete step by step answer:
Here we have been provided with the velocity of a particle as a function of time as $v=2t+1$. We have been asked to find the position x at any instant of time t given that initially the particle is at origin.
Now, we have to determine the position x that means the displacement along x. We know that the velocity of a particle along a direction is the derivative of displacement with respect to time along that direction with respect. So the given velocity must be along the x direction.
$\Rightarrow v=\dfrac{dx}{dt}$
So replacing v in $v=2t+1$ we get,
$\Rightarrow \dfrac{dx}{dt}=2t+1$
Multiplying both the sides with $dt$ we get,
$\Rightarrow dx=\left( 2t+1 \right)dt$
Integrating both the sides we get,
$\Rightarrow \int{dx}=\int{\left( 2t+1 \right)dt}$
Now we need to consider proper limits. It is given that the particle starts from origin that means at t = 0 we have x =0. Also, we are finding the position x at any instant t so we have the limits as:
$\begin{align}
  & \Rightarrow \int_{0}^{x}{dx}=\int_{0}^{t}{\left( 2t+1 \right)dt} \\
 & \Rightarrow \left[ x \right]_{0}^{x}=\left[ \dfrac{2{{t}^{2}}}{2}+t \right]_{0}^{t} \\
 & \Rightarrow \left[ x \right]_{0}^{x}=\left[ {{t}^{2}}+t \right]_{0}^{t} \\
\end{align}$
Substituting the limits we get,
 \[\begin{align}
  & \Rightarrow \left[ x-0 \right]=\left[ \left( {{t}^{2}}+t \right)-\left( 0+0 \right) \right]_{0}^{t} \\
 & \Rightarrow x=\left( {{t}^{2}}+t \right) \\
\end{align}\]

So, the correct answer is “Option a”.

Note: Note that we use these formulas regarding the displacement, velocity and acceleration in the topic ‘Application of Derivatives’ also these things appear in the chapter ‘Kinematics’ in physics. Remember that the derivative of velocity with respect to time is known as the acceleration. You must be careful while determining the upper and lower limits.