Question

A particle moves on a straight line and its velocity as $v=(3-t)m/s$. Find the distance travelled by it in the first 5 seconds.
A) 2.5m
B) 4.5m
C) 6.5m
D) 8m

Answer 1

Hint: The term ‘velocity’ is the unit to measure the displacement of an object traveled within a certain time. We know that velocity is a vector quantity. Hence it has both magnitude and direction. The displacement can be negative but the time cannot be negative. The term ‘distance’ is also similar to the displacement but the difference is that distance does not involve the direction. Therefore, we can say that the distance can be higher than the displacement if the particle travels in one direction.

Formula used:
$v={\displaystyle \frac{dx}{dt}}$
$\Rightarrow dx=vdt$
Where, $v$=velocity, and ${\displaystyle \frac{dx}{dt}}$=distance changes with time $t$.

Complete step by step answer:
Given, $v=(3-t)m/s$.
Velocity will be positive only when t is up to 3 seconds. If $t$ increases, velocity becomes negative. It denotes that the particle moves in the opposite direction after 3 seconds. Hence, we calculate the distance x separately for up to 3 seconds and 3 to 5 seconds.
(i) The distance traveled by the particle up to 3 seconds,
$\underset{0}{\overset{3}{\int }}dx=\underset{0}{\overset{3}{\int }}\left(3-t\right)dt$
$x={\left[3t-{\displaystyle \frac{t^2}{2}}\right]}_0^3$
$x=9-{\displaystyle \frac{9}{2}}$
$\Rightarrow x=4.5m$…………(1)
(ii) The distance traveled by the particle within the time period of 3 to 5 seconds,
$\underset{3}{\overset{5}{\int }}dx=\underset{3}{\overset{5}{\int }}\left(3-t\right)dt$
$x={\left[3t-{\displaystyle \frac{t^2}{2}}\right]}_3^5$
$x=15-{\displaystyle \frac{25}{2}}-9+{\displaystyle \frac{9}{2}}$
$\Rightarrow x=-2m$(Distance cannot be negative)
Hence, $x=2m$………….(2)
Adding (1) and (2) gives us the total distance traveled by the particle.
(1)+ (2) equal to, $x=4.5+2$
$\Rightarrow x=6.5m$

Therefore, the particle traveled a distance of 6.5m in a straight line in 5 seconds. Hence the correct option is C.

Additional information:
(i) Velocity is the rate of distance traveled. It is the vector quantity. It has both magnitude and direction. Velocity is different from speed as speed is the scalar quantity. A scalar quantity has only magnitude no direction.
(ii) The SI unit of velocity is $m{s}^{-1}$. And the dimension of velocity is $L{T}^{-1}$.
(iii) The rate of velocity is called acceleration. Acceleration is the ratio of velocity and time. Hence its unit is $m{s}^{-2}$ and the dimension of acceleration is $L{T}^{-2}$.

Note:
If the particle travels without any change in direction, we can do a single integration using a required limit. Because if direction changes within a given time limit, the velocity will be the algebraic sum of the total distance traveled. Hence we could not find the total distance traveled by a particle by single integration in the case of direction change.