Question

The distance travelled by a particle starting from rest and moving with an acceleration $\dfrac{4}{3}$ ms$^{-2}$ in the third second is-
(A) $\dfrac{10}{3}$ m
(B) $\dfrac{19}{3}$ m
(C) 6 m
(D) 4 m

Answer 1

We are given the constant acceleration that the particle undergoes. The point to be noted here is that we are asked to find the distance covered in the third (or nth) second and not distance covered upto the third second of the particle motion.

Complete answer:
Let us start by writing the acceleration in terms of Differential equation as:
$a = \dfrac{d^2 x}{dt^2}$.
Now, upon integrating this equation twice we get:
$x = a \dfrac{t^2}{2}$.
Here, we have taken as constant as has been given in the question. We have neglected the constants of integration here because we will see that since particles start from rest, we will get them as zero.
Now, since we will need to determine the distance travelled in between 2nd and 3rd second, we put the limits as:
$d = a \left( \dfrac{t^2}{2} \right) \Bigg|_2^3$
where d has been used to denote the total distance travelled.
Therefore, upon keeping the values, we get:
$d = \dfrac{4}{3} \left( \dfrac{(3^2 - 2^2)}{2} \right) \Bigg|_2^3 = \dfrac{10}{3}$m.

Therefore, the correct answer is option (A).

Additional information:
Differential equations help in solving the questions based on motion very easily. If we denote the position by x and we differentiate it with respect to time once, we find the velocity. If we differentiate the velocity with respect to time we get the acceleration for the motion. For the case of rest, or constant motion, the acceleration is zero i.e., it does not change with time.

Note:
While writing the differential equation here, directly a has been equated with its differential equivalent i.e., second order time derivative of position. While handling the problem this way does not require for one to remember any type of specific formula it also eases up the calculations. One has to be careful here and check if any constants appear after any integration. The constants of integration entirely depend upon the initial conditions. Therefore, one has to check the value of constants after performing each integration by putting initial conditions.