+91-8076753736 [email protected]

-->

Motion In Straight Line

Question

Wired Faculty App

The velocity of a particle at any instant is given by v = 5t + 3. From the v- t graph, find the displacement of the particle between t = 2 to t = 5.

Solution

Here,
Velocity of a particle at any instant, v = 5t + 3
So,
Velocity|_t=2 is v(2) = 13 m/s
velocity|_t=5 is v(5) = 28 m/s
We know displacement of particle is equal to area under v-t graph.
$space space therefore$ Distance travelled = Area of trapezium ABCD
$space equals space 1 half left parenthesis AB plus BC right parenthesis cross times DC space equals 1 half left parenthesis 13 plus 28 right parenthesis cross times 3 equals 61.5 straight m$

Some More Questions From Motion in Straight Line Chapter

What is the most essential thing to study the motion of a body?

Can a body be always at rest in some frame?

Can a body be in motion in the frame associated with the body itself?

When an object in motion can be considered as particle or point object?

Can you think of physical phenomena involving the moon in which moon cannot be treated as particle?

A train is crossing the bridge. Can it be taken as a particle?

Can a train be taken a particle? Give an example in support of your answer.

A particle is moving along curved line in space. Is the motion one dimensional?

State which of the following are examples of one, two or three dimensional motion:

(a) A kite flying in the sky.

(b) The earth revolving around the sun.

(c) A train moving along the equator of the earth in clockwise direction.

(d) Accelerated motion of a particle in a straight line.

Define distance.

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation