Question

If \[x = 5t + 3{t^2}\] and \[y = 4t\]are the x and y coordinates of a particle at any time t seconds where x and y are in metres, then the acceleration of the particle
A. is zero throughout its motion
B. is a constant throughout its motion
C. depends only on its y component
D. varies along both x and y direction

Answer 1

Hint: The above problem can be resolved using the concept of acceleration of the particle travelling along the x-axis as well as the y-axis. The acceleration in the respective directions can be calculated using the double differentiation of the mathematical form of distance travelled in both the axis, Then, the result is analysed to undertake the conclusion from the available options.

Complete step by step answer:
The distance travelled in x axis is,
\[x = 5t + 3{t^2}................................\left( 1 \right)\]
 The distance travelled in y axis is,
\[y = 4t...............................\left( 2 \right)\]
Differentiating the equation 1 and 2 twice to obtain the acceleration along x and y direction as,
\[\begin{array}{l}
{a_x} = \dfrac{{{d^2}\left( {5t + 3{t^2}} \right)}}{{dt}}\\
{a_x} = 6........................\left( 3 \right)
\end{array}\]
 Similarly along the y-axis,
\[{a_y} = 0...................\left( 4 \right)\]
From the equation 3 and 4, it is clear that acceleration of the particle along the x- axis and y- axis is having different magnitudes.
Therefore, the acceleration of the particle varies along both x and y direction

So, the correct answer is “Option D”.

Note:
To resolve the given problem, remember the fact that the acceleration can be calculated using the differentiation of the expression of the velocity. And the velocity can be calculated by differentiating the expression of the distance travelled by the object in either direction. These mathematical relations provide the appropriate results to calculate the acceleration in either direction. Moreover, the kinematic relations for the acceleration and the resultant acceleration can also be used to conclude the given problem.