Question

It has been given that the position vector of a particle varies with time according to the relation $15t^2\hat{i}+(4-20t^2)\hat{j}$​. Find out the magnitude of the acceleration at $t=1$?
$\begin{array}{rl}& A.40\\ & B.100\\ & C.25\\ & D.50\end{array}$

Answer 1

Hint: First of all take the derivative of the position vector given. This will be the velocity of the vector. Differentiate the velocity vector which will give the acceleration vector. Substitute the values needed and then find the resultant acceleration. This will help you in answering this question.

Complete step by step answer:
It has been mentioned in the question that the position vector of the particle is,
$\overrightarrow{r}\left(t\right)=15t^2\hat{i}+(4-20t^2)\hat{j}$
The derivative of this function will give us the velocity vector needed. Therefore we can differentiate this equation with respect to time. This can be written as,
${\displaystyle \frac{\partial }{\partial t}}\overrightarrow{r}\left(t\right)={\displaystyle \frac{\partial }{\partial t}}(15t^2\hat{i}+(4-20t^2)\hat{j})$
Perform the differentiation with respect to time.
$\begin{array}{rl}& \overrightarrow{v}\left(t\right)=(15\times 2t\hat{i}+(0-20\times 2t)\hat{j})\\ & \Rightarrow \overrightarrow{v}\left(t\right)=(30t\hat{i}+(0-40t)\hat{j})\end{array}$
The acceleration is given as the derivative of the velocity. Hence we have to take the derivative of the velocity vector with respect to time taken in order to calculate the acceleration of the particle.
This can be shown as,
${\displaystyle \frac{\partial }{\partial t}}\overrightarrow{v}\left(t\right)={\displaystyle \frac{\partial }{\partial t}}(30t\hat{i}+(-40t)\hat{j})$
Let us perform the differentiation with respect to the time taken. That is,
$\begin{array}{rl}& \overrightarrow{a}\left(t\right)={\displaystyle \frac{\partial }{\partial t}}(30t\hat{i}+(-40t)\hat{j})\\ & \overrightarrow{a}\left(t\right)=30\hat{i}-40\hat{j}\end{array}$
The resultant acceleration can be found out by taking the resultant of this vector. There we can write that,
$$\overrightarrow{a}=\sqrt{{30}^2+{(-40)}^2}$$
This will be simplified as,
$$\left|\overrightarrow{a}\right|=50$$
Therefore the acceleration of the body is found out.
The correct answer is given as option D.

Note:
A position or position vector is otherwise referred to as the location vector or the radius vector. It is a Euclidean vector that indicates the position of a point A in space. This is with respect to any randomly taken reference. The position vector will change in length or in direction or in both length and direction as the point is moving.