Question

The position $x$ of a particle at time \[t = 0\] is \[x = - 15{\text{ }}m\]. If the particle starts moving with a constant speed of \[6{\text{ }}m/s\], then the position of particle after 3 seconds is (Direction of motion of the particle is positive $x$)
A) $x\, = \,18\,m$
B) $x\, = \, - 3\,m$
C) $x\, = \,3\,m$
D) $x\, = \,12\,m$

Answer 1

Hint : The direction of motion will decide the displacement of the particle from its mean position. We will use the laws of motion to determine the position of the particle after 3 seconds.

Formula used: In this question, we will use the following formula:
$d = vt$ where $d$ is the distance travelled by the particle travelling with velocity $v$ and time $t$

Complete step by step answer
We’ve been given the position of the particle at time \[t{\text{ }} = {\text{ }}0\] is \[x = - 15{\text{ }}m\]. Then the particle starts moving in the direction of the positive x-axis with a constant speed of \[6{\text{ }}m/s\]. As a result, the distance covered by the particle will be a product of the velocity of the particle with the time it travels for which can be mathematically written as
$d = vt$
Now the distance travelled by the particle when it moves for 3 seconds with a speed of \[6{\text{ }}m/s\]will be,
$d = 6 \times 3$
$ \Rightarrow d = \,18\,m$
This is the displacement of the particle from its initial position in the direction of the positive $x$-axis. Since it was originally at the coordinate \[x = - 15{\text{ }}m\], the position of the particle after 3 seconds will be
$x = - 15 + 18$
$ \Rightarrow x = + 3\,$
Hence the position of the particle after 3 seconds will be $( + 3)$ which corresponds to option (C).

Note
We must be careful in taking into account the direction of the velocity of the particle when calculating the final coordinate of the particle. Had the particle been moving in the negative $x$-direction, its coordinate would shift to the left side but since the velocity is in the positive $x$-direction, it will move to the right on the $x$-axis.