Question

If the velocity of a particle is \[v = At + B{t^2}\] , where A and B are constants, then the distance travelled by it between 1 s and 2 s is
A. \[\dfrac{3}{2}A + 4B\]
B. \[3A + 7B\]
C. \[\dfrac{3}{2}A + \dfrac{7}{3}B\]
D. \[\dfrac{A}{2} + \dfrac{B}{3}\]

Answer 1

Hint: The above problem can be resolved using the concepts of the velocity and the distance covered by the particles any specific instant of time. In the given problem, the velocity of the particles is given in the form of a quadratic equation, and it is known that on applying the differentiation of that equation, the final value of the distance covered for the given interval of time can be calculated. The speed and distance covered by any object depends on the magnitude as well as the direction. Moreover, the other kinematic relations can also be taken into account involving the appropriate method of calculating the values of variables like speed, distance, displacement, acceleration and many more.

Complete step by step answer:
Given:
The velocity of the particle is, \[v = At + B{t^2}\] .
The distance travelled by the particle is given as,
\[\begin{array}{l}
S = \int\limits_{t = 1}^{t = 2} v dt\\
S = \int\limits_{t = 1}^{t = 2} {\left( {At + B{t^2}} \right)} dt\\
S = \int\limits_{t = 1}^{t = 2} {\left( {At} \right)} dt + \int\limits_{t = 1}^{t = 2} {\left( {B{t^2}} \right)} dt\\
S = A\left( {\dfrac{{{t^2}}}{2}} \right)_{t = 1}^{t = 2} + B\left( {\dfrac{{{t^3}}}{3}} \right)_{t = 1}^{t = 2}
\end{array}\]
Further solving as,
\[\begin{array}{l}
S = A\left( {\dfrac{{{{\left( 2 \right)}^2} - {{\left( 1 \right)}^2}}}{2}} \right) + B\left( {\dfrac{{{{\left( 2 \right)}^3} - {{\left( 1 \right)}^3}}}{3}} \right)\\
S = \dfrac{3}{2}A + \dfrac{7}{3}B
\end{array}\]
Therefore, the distance covered by the particle is \[\dfrac{3}{2}A + \dfrac{7}{3}B\]

So, the correct answer is “Option C”.

Note:
To resolve the given condition, one must be aware of the fact that the kinematic variables like distance, speed and acceleration are related to each other both mathematically as well as analytically. To obtain the magnitude of the velocity, out of the given data regarding the distance covered by an object, one can apply the method of single differentiation. Similarly, to find the acceleration, the double differentiation of the distance can be applied. Moreover, these variables are used to resolve the typical kinematic conditions involving the motion in one-dimension, two-dimension and in three dimensions.