Question

The following velocity–time graph represents a particle moving in the positive x–direction. Analyse its motion from 0 to 7 s. Calculate the displacement covered and distance travelled by the particle from 0 to 2 s.

Answer 1

I completed the graph by providing the time t = 1.75s at which velocity becomes zero for the first time as shown in pic.


Final Answer :
Distance = 7/8 m
Displacement = 5/8m

Steps:
1) Analyzing the Motion from t =0 to t = 7s:
Initially, the speed of particle increases to its maximum value, then again it's value decreases to 0.
=> After that value of speed again increases but in opposite direction, then it remains constant.
After acquiring constant speed, it decreases to 0 and remains 0.

2) Second Part :
Distance covered from t = 0 to t = 2s.
= | Displacement from t = 0 to t = 1.75 s) + | Displacement from t = 1.75 s to t = 2s|
= |Area under v-t in required time interval |
+ | Area under v-t in required time interval ¦
=
$\frac{1}{2} \times 1.75 \times 2 + \frac{1}{2} \times (2 - 1.75) \times 1 \\ = \frac{6}{8} + \frac{1}{8} \\ = \frac{7}{8} m$


Displacement = Area under v-t graph from t = 0 to t = 1.75 s - Area from t = 1.75s to t = 2s

$= \frac{1}{2} \times 1.75 \times 2 - \frac{1}{2} \times (2 - 1.75) \times 1 \\ \\ = \frac{6}{8} - \frac{1}{8} = \frac{5}{8} m$

Answer 2

Answer:

distance travelled by the particle = 1.5m

displacement covered = -1.5m

Explanation:

It is a uniform motion.

OAB

Distance travelled by the particle = area of triangle OAB = 1/2*b*h

OB = b, PA = h

1/2*1.5*2 = 1.5m

Displacement covered = 1/2*b*h

1/2*1.5*(-2)

= -1.5m

A particle moves with -ve velocity along OA, then moves with uniform velocity and retards uniformly then it comes to rest at E.