(1) A particle starts from rest at t = 0 and moves in a straight line with an acceleration as shown below. The velocity of the particle at t = 3 s is
(a) 2 ms–1
(b) 4 ms–1
(c) 6 ms–1
(d) 8 ms–1
According to the acceleration – time graph shown, the particle has an acceleration of 4 ms–2 during the first two seconds. Therefore, its velocity (v2) at the end of 2 seconds is given by
v2 = v0 + at = 0 + 4×2 = 8 ms–1
From 2 second to 3 second the particle has a retardation of 4 ms–2. Hence its velocity (v3) at the end of 3 seconds is given by
v3 = v2 – at = 8 – 4×1 = 4 ms–1 [Option (b)].
(2) Two cars A and B are moving with same speed of 45 km/hr along same direction. If a third car C coming from the opposite direction with a speed of 36 km/hr meets two cars in an interval of 5 minutes, the distance of separation of two cars A and B should be (in km)
(a) 6.75
(b) 7.25
(c) 5.55
(d) 8.35
(e) 4.75
The relative velocity of car A with respect to car B (and that of car B with respect to A) is zero since they have the same speed in the same direction. The relative velocity of car C with respect to A and B is 45 + 36 = 81 km/hr. Since the car C takes a time of 5 minutes to cover the distance between A and B, the separation between A and B is 81×(5/60) km = 6.75 km.
(3) An object is dropped from rest. Its v - t graph is
The correct option is (a) since the velocity ‘v’ is directly proportional to the time t in accordance with the equation v = v0 + at where v0 is the initial velocity which is zero and a is the acceleration which is the constant acceleration due to gravity. The graph should evidently pass through the origin.
Suppose you were asked to draw the v – t graph in the case of a ball projected vertically upwards with a velocity u. If you are asked to draw the graph from the instant the ball leaves your hand to the instant you catch it while returning, you can do it as shown, ignoring the air resistance and the time taken for the velocity to reduce to zero on hitting your hand.
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