An object moving with a speed of 6.25m/s, is decelerated at a rate given by dv/dt =- 2.5√v, where v is the instantaneous speed. The time taken by the object, to come to rest, would be
2 s
4 s
8 s
8 s
A.
2 s
A particle of mass m is at rest at the origin at time t = 0. It is subjected to a force F (t) = F0e–bt in the x direction. Its speed v(t) is depicted by which of the following curves?
C.
A particle is moving with velocity, where K is a constant. The general equation for its path is
y = x2 + constant
y2 = x + constant
xy = constant
xy = constant
D.
xy = constant
The velocity of the particle,
A boy can throw a stone up to a maximum height of 10 m. The maximum horizontal distance that the boy can throw the same stone up to will be
10 m
D.
mMaximum speed with which the boy can throw stone is
Range is maximum when projectile is thrown at angle of 45o, Thus,
The figure shows the position –time (x – t) graph of one–dimensional motion of a body of mass 0.4 kg. The magnitude of each impulse is
0.4 (Ns)
0.8 Ns
1.6 Ns
1.6 Ns
B.
0.8 Ns
From the graph it is a straight line, so uniform motion. Because of impulse direction of velocity changes as can be seen from the slope of the graph.
(∴ impulse (l) = f x t
⇒ I = mat
Initial velocity, v1 = 2/2 = 1 ms-1
Final velocity v2 = 2/2 = - 1 ms-1
pi = mv1 = 0.4 N-s
pf = mv2 = -0.4 N-s
J = pf-pi = -0.4-0.4
= - 0.8 N-s
|J| = 0.8 N-s