Question

The x and y coordinates of a particle at any time t are given by $x = 7t + 4{t^2}$ and $y = \sqrt {95t} $, where x and y are in metre and t in seconds. The rate of change of speed of particle at t = 5 sec is:
(A) $\dfrac{{47}}{6}m/{s^2}$
(B) $8m/{s^2}$
(C) $20m/{s^2}$
(D) $40m/{s^2}$

Answer 1

Hint: To answer this question we should first begin the answer by differentiating the given expression. Then we have to find the value of ${a_1}$. Once we get that we have to consider the y expression and find out the value of a at the end. For finding the value of a we need to consider the value of ${a_x}$ and ${a_y}$ and find the rate of change of speed of the particle. This will give the answer to the required question.

Complete step by step answer:
We should know that:
$x = 7t + 4{t^2} {v_x} = \dfrac{{dx}}{{dy}} = 7 + 4 \times 2t$
Now we can write that the value of ${a_1}$.
So the value of ${a_1}$ is here:
${a_1} = \dfrac{{d{v_x}}}{{dt}} = 8m/{\sec ^2}$
Now for the expression of y is given by:
$
  y = 5t \\
   \Rightarrow {v_y} = \dfrac{{dy}}{{dt}} = 5 \\
 $
And the value of ${a_y}$= 0.
So the value of a can be given as:
$a = \sqrt {a_x^2 + a_y^2} = \sqrt {{{(8)}^2} + 0} = 8m/{s^2}$
The value of a does not depend on time.
So we can say that the rate of change of speed of particles at t = 5 sec is $8m/{s^2}$.

 So the correct answer is option B.

Note: For a graphical method we should remember that the horizontal axis is known as the x axis. And the vertical axis is known as the y axis. Every point on the graph is specified with an ordered pair of numbers, which will be consisting of numbers from both the coordinates, that is the x coordinate and the y coordinate.
In case of solving problems which involve a quantity with values from both the coordinates (x and y) we need to find the actual value, from the square root of both the coordinates square.