Question

A particle is moving along x-axis. It $x-t$ graph is given below-
Find:
(i) Displacement in $8sec$
(ii) Distance in $8sec$
(iii) Average velocity in $8sec$
(iv) Velocity at $t=1sec$

Answer 1

(i) Displacement in $8sec$

$\text{Displacement}=\text{final position}-\text{initial position}$

From graph-

$\text{Final position}=0m$; at 8 sec

$\text{Initial position}=20m$; 0 sec

$\Rightarrow \text{Displacement}=\text{final position}-\text{initial position}$

$\Rightarrow \text{Displacement}=\text{0}-\text{20}$

$\Rightarrow \text{Displacement}=-20m$

(ii) Distance in $8sec$

$\text{Distance}=\text{Total length of the path}$

From graph-

Distance from $0s$ to $2s$

$d_1=20m$

Distance from $2s$ to $4s$

$d_2=0m$; as potion of particle remains same

Distance from $4s$ to $8s$

$d_3=40m$

total distance-

$d=d_1+d_2+d_3$

$\Rightarrow d=20+0+40$

$\Rightarrow d=60m$

(iii) Average velocity$\left(V_{avg}\right)$ in $8s$-

$V_{avg}=\frac{\text{total displacement}}{\text{total time}}$

$\Rightarrow V_{avg}=\frac{-20}{8}$

$\Rightarrow V_{avg}=-2.5m{s}^{-1}$

(iv) Velocity at $t=1s$

This is instant velocity which is equal to the slope of the graph at that point.

$V_{t=1}=\frac{\text{perpendicular}}{\text{base}}$

$\Rightarrow V_{t=1}=\frac{40-20}{2-0}$

$\Rightarrow V_{t=1}=\frac{20}{2}$

$\Rightarrow V_{t=1}=10m{s}^{-1}$