Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Describe the motion as shown by the following velocity time graphs.
A)

B)

seo-qna
Last updated date: 17th Apr 2024
Total views: 35.1k
Views today: 1.35k
Answer
VerifiedVerified
35.1k+ views
Hint: The slope of the velocity-time graph gives the acceleration of the body. When the line describing motion of the particle is parallel to the X-axis, the velocity of the particle is considered to be 0. The motion of the body can be defined using acceleration, velocity and the displacement of the body. The slope of any velocity-time graph is the acceleration and the area of the graph is the displacement.

Complete step by step answer:
(A)

As we can see from the above figure, the point A is in the positive Y axis that means, the velocity of the particle is positive.
We know, Slope of the velocity-time graph gives us the acceleration. In the region AB, we see a slope which is decreasing, therefore, we can say the acceleration is in the negative direction, also termed as deceleration.When the line describing motion of the particle is parallel to the X-axis, the velocity of the particle is considered to be 0. Thus, the region BC is constant zero velocity.

Further, in region CD, we observe a negative slope, that means the body again decelerates at a constant rate.

(B)

In the above figure, we can observe the following:
Region AB has a constant velocity of some magnitude, as the line of motion is parallel to the x axis.
Region BC, it is seen that the slope of the curve proceeds towards the negative axis, as it has a negative slope. Thus the body decelerates.
In the region CD, the body continues moving further towards the negative axis, due to constant deceleration.

Note: When velocity of a particle is 0, the curve in velocity-time graph coincides with the X axis. Acceleration refers to the increase in speed of a body with respect to time and deceleration refers to decrease in speed of a body with respect to time. Deceleration is also referred to as negative acceleration.