1. An object is moving along a straight line with uniform acceleration. The following table gives the velocity of the object at various instants of time. Time (s) 0 1 2 3 4 5 6 Velocity (m/s) 2 4 6 8 10 12 14 Plot the graph and calculate,’ i. Find the velocity of the object at the end of 2.5s. ii. Calculate the acceleration. iii. Calculate the distance covered in the last 4 seconds.
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Answer:Following is the graph corresponding to the uniformly accelerated motion
a) Velcoity at 2.5 seconds is 7m/s
b) Acceleration = slope of velocity time graph
=
(6−0)
(14−2)
=2m/s
2
c) Distance covered in last 4 seconds is the area under the curve for last 4 second
i.e.
Total distance covered in last 4 seconds
= area under ABCD
= area of ABED + area of BCE
=((6−0)×(6−2))+
2
(14−6)×(6−2)
=24+16
=40m
Explanation:
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