03. acceleration along a straight line done
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The displacement of a particle along a straight line at time t is given by x = a0-a1t + a2t2. The acceleration of the particle isA. a0B. a1C. 2a2D. a2
ANSWER: D
The acceleration of a moving body can be found fromA. area under velocity -time graphB. area under distance -time graphC. slope ofthe velocity -time graphD. slope of the distance -time graph
ANSWER: C
137 A lift ascends with an upward acceleration of 0.2 m/s2. A loose bolt dropped from the ceiling of the lift which is at a height 5 m from the floor. The timetaken by the bolt to strike the floor (g = 9.8 m/s2)A. 1.0sB. 1.2sC. 1.5s
D. 1.7sANSWER: A
138 A pebble thrown vertically upwards from a bridge with an initial velocity of 9.8 m/s strikes the water after 4 seconds. The height of the bridge is (g = 9.8 m/s2)A. 9.8mB. 39.2mC. 14.7mD. 19.6m
ANSWER: B
A vertical circular disc has different grooves along various chords as shown. Particles are released from the upper end '0'. The times taken by the particles to reach the ends of the grooves OA, DB, DC are respectively in the ratio.1. What is the time taken by the particle to move along OA [ANS: t1 = sqrt(2D/g)]2. What is the time taken by the particle to move along OA [ANS: t1 = sqrt(2D/g)]A. 1:sqrt(3):2B. 1:2:sqrt(3)C. 1:2:3D. 1:1:1
ANSWER: A
A.B.C.D.
ANSWER: A
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A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.
D.ANSWER: A
A.B.C.D.
ANSWER: A
A.B.
C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.
B.C.D.
ANSWER: A
A.B.C.D.
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ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.
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B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.D.
ANSWER: A
A.B.C.
D.ANSWER: A
A.B.C.D.
ANSWER: A