10. Study the given graph and answer the following question
a. Which part of the graph shows accelerated motion?
b. Which part of the graph shows retarded motion?
c. Calculate the distance travelled by the body in first 3 seconds of
journey graphically.
d. Calculate the acceleration of the body between 6 s and 7s.
Answers
Answer:
Given graph is a VELOCITY-TIME graph of an object.
In VT graph:
1. The slope of the line at a particular time gives the acceleration of the object at that time.
2. Area under the graph gives the displacement of the object.
_______________________________
Coming to the questions:
1. Part AB is accelerated motion as the slope of the graph(which gives the acceleration) is positive.
acceleration = slope
slope = \frac{(4 - 0) \: m/s}{(4 - 0)s} = \frac{4}{4} = 1 \ m/s^2slope=
(4−0)s
(4−0)m/s
=
4
4
=1 m/s
2
So acceleration = 1 m/s^2
2.Part CD is retarded motion as the slope of the graph(which gives the acceleration) is negative.
slope = \frac{(0 - 4) \: m/s}{(14 - 10)s} = \frac{ - 4}{4} = - 1 \ m/s^2slope=
(14−10)s
(0−4)m/s
=
4
−4
=−1 m/s
2
So acceleration = -1 m/s^2
Thus retardation is 1 m/s^2
3.
distance of the body in first 4 seconds = area under graph from t=0 to t=4s
area under graph from t=0 to t=4s = area of ∆ABE = 1/2 × 4 × 4 = 8m
Thus distance of the body in first 4 seconds is 8m