# H. C. Verma Solutions – Chapter 3, Question 8

H. C. Verma, Chapter 3, Question 8 The figure shows the graph of velocity versus
time for a particle going along the x-axis. Find (a) the acceleration, (b) the distance
travelled in 0 to 10 seconds and (c) the displacement in 0 to 10 seconds. Acceleration is given by change in velocity
divided by time. As the final velocity is 8 meter per second and initial velocity is
2 meter per second, the change in velocity is 6 meters per second. Total time taken is
10 seconds. Therefore, acceleration is equal to 6 by 10 or 0.6 meter per second square.
This is answer to part A of the question. In a velocity time graph, displacement is
equal to the area under the graph. Therefore, displacement in this case is area of the shaded
region. The area of trapezium is half into sum of parallel sides into distance between
the parallel sides. Therefore, displacement in this case is half into 2 plus 8 into 10.
That is 50 meters. Since the velocity of the particle does not change its direction during
the motion, the distance and displacement is equal. Therefore, both the displacement
and distance is 50 meters. This is answer to part b and part c of the question. In order to request solution to any question
from H. C. Verma book, post the question number and chapter number in comments below. LIke
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## 7 thoughts on “H. C. Verma Solutions – Chapter 3, Question 8”

1. Raman Kumar says:

Sir hcv question no. 32 of chapter 10

2. Sat-tech In says:

thank you

3. Nari Salih says:

thank you sir

4. Tupil Video says:

Sir,l wasn't class 9 physics HC verma chapter 2 solution by 2nd April

5. Tupil Video says:

l wan't u to upload hc verma dil of all chapter of class 9

6. 52 magiks says:

Sir do you give solutions only for HC Verma or any other book.

7. Anshu kumar says:

Nice