Question

The position of a particle vary \[x = 4t - 2{t^2}.\] The distance covered by the particle in 4 s is
A. 5m
B. 1m
C. 2m
D. 4m

Answer 1

Hint: The position of the particle that is given is the displacement of the particle. The displacement of the particle can be defined as the shortest distance it travels along a path. The distance of the particle can be said as the total amount travelled by it along the path. Since, velocity is the term that relates the displacement and time hence it is defined as the rate of change of displacement.

Complete step by step solution:
The displacement of the particle is given as the position of the particle in the question. We know that the displacement is the shortest distance travelled by the particle in the path. It is also a vector quantity and hence it depends on the direction of motion.

The distance of the particle can be defined as the total movement covered by the particle along the path. The distance does not depend on the direction of the movement of the particle. Both the distance and displacement have the same SI unit as ‘m’.

Consider the given position as the displacement of the particle, so,
\[x = 4t - 2{t^2}\]
By differentiating the above equation, we can get the velocity of the particle,
\[\dfrac{{dx}}{{dt}} = 4 - 2t\]
\[\Rightarrow \dfrac{{dx}}{{dt}} = 4(1 - t)\]
At t =1s, we get,
\[v = 0\]
At t = 4s, we get,
\[v = - 12\,m{s^{ - 1}}\]
So the velocity is in the negative direction. Then the distance at t= 1s is,
\[x = 4(1) - 2{(1)^2}\]
\[\Rightarrow x = 2m\]
Since v = 0m/s then the particle travels in negative direction, so at t = 2s,
\[x = 4(2) - 2{(2)^2}\]
\[\Rightarrow x = 8 - 8 = 0\]
At t= 2 s the distance x is 0 metre, which means the particle travels back, hence when t = 4s the distance is
\[x = 2 + 2\]
\[\therefore x = 4m\]

So, the correct answer is option D.

Note : Here the particle has velocity zero at t = 1s, which means the motion stops at 1s. So the distance covered at \[{1^{st}}\] is 2m and at \[{2^{nd}}\] second is zero metre. Similarly at the \[{3^{rd}}\] second, it will travel to a distance of 2m and at \[{4^{th}}\] second it will be zero again. So the total distance travelled at the end of 4s is 4m.