A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration in this time

Q:A car is moving eastwards with velocity 10 m/s. In 20 sec, the velocity changes to 10 m/s northwards. The average acceleration in this time:

(A)1/√2 m s-2 towards N-W

(B)1/√2 ms-2 towards N-E

(C) 1/2 m s-2 towards N-W

(D) 1/2 ms-2 towards N

Solution: The change in velocity is

Δv = v2 – v1
= 10j – 10i
Δv has magnitude = 10√2 and directed due  N – W
Average acceleration =Δv ⁄Δt =10√2/20
= 1/√2 ms-2 and directed due N-W

Ans: (A)

A boy standing, at the top of a tower of 20 m height drops a stone. Assuming, g = 10 ms–2, the velocity with which it hits the ground is

Question: A boy standing, at the top of a tower of 20 m height drops a stone. Assuming, g = 10 ms–2, the velocity with which it hits the ground is
(a) 20 m/s
(b) 40 m/S
(c) 5 m/s
(d) 10 m/s

Ans: (a)

The motion of a particle along a straight line is described by equation x = 8 + 12t – t3 where, x is in metre and t in sec. The retardation of the particle when its velocity becomes zero, is

Q:The motion of a particle along a straight line is described by equation
x = 8 + 12t – t3
where, x is in metre and t in sec. The retardation of the particle when its velocity becomes zero, is
(a) 24 ms–2
(b) zero
(c) 6 ms–2
(d) 12 ms–2

Ans: (d)

A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5s, the next 5s and the next 5s respectively. The relation between h1, h2 and h3 is

Q:A stone falls freely under gravity. It covers distances h1, h2 and h3 in the first 5s, the next 5s and the next 5s respectively. The relation between h1, h2 and h3 is
(a) h1 = 2h2 = 3h3
(b) h1 = h2/3 = h3/5
(c) h2 = 3h1 and h3 = 3h2
(d) h1 = h2 = h3

Ans: (b)