The dependence of velocity of a body with time is given by the equation v=20+0.1t².The body is in

uniform retardation
uniform acceleration
non-uniform acceleration
zero acceleration.

Solution

On differentiating, acceleration = 0.2t ⇒ a = f (t)

Which one of the following equations represents the motion of a body with finite constant acceleration ? In the see quations, y denotes the displacement of the body at time t and a, b and c are constants of motion.

y = at
y=at+bt²
y=at+bt²+ct³
y=^a⁄_t+bt

Solution

y α t²;v - α t';a α t°

The displacement x of a particle varies with time t as x = ae^-αt+ be^βt, where a, b, α and β are positive constants.The velocity of the particle will

be independent of α and β
drop to zero when α = β
go on decreasing with time
go on increasing with time

Solution

In the given figure the distance PQ is constant.SQ is a vertical line passing through point R.A particle is kept at Rand the plane PR is such that angle θ can be varied such that R lies on line SQ. The time taken by particle to comedown varies, as the θ increases

decreases continuously
increases
increases then decreases
decreases then increases

Solution

The acceleration due to gravity on planet A is nine times the acceleration due to gravity on planet B.A man jumps to a height 2m on the surface of A. What is height of jump by same person on planet B?

2/3 m
2/9 m
18 m
6 m

Solution

Since the initial velocity of jump is same on both planets

Two cars A and B are travelling in the same direction with velocities v_A and v_B(v_A > v_B). When the car A is at a distanced behind the car B the driver of the car A applies brakes producing a uniform retardation a. There will be no collision when

d < \(\frac{(V_{A}-V_{B})^{2}}{2a}\)
d < \(\frac{V_{A}^{2}-V_{B}^{2}}{2a}\)
d < \(\frac{\left (V_{A}-V_{B} \right )^{2}}{2a}\)
d > \(\frac{V_{A}^{2}-V_{B}^{2}}{2a}\)

Solution

Let A, B, C,D be points on a vertical line such that AB = BC = CD. If a body is released from position A, the times of descent through AB, BC and CD are in the ratio.

1:\(\sqrt{3}\)-\(\sqrt{2}\):\(\sqrt{3}\)+\(\sqrt{2}\)
1:\(\sqrt{2}\)-1:\(\sqrt{3}\)-\(\sqrt{2}\)
1:\(\sqrt{2}\)-1:\(\sqrt{3}\)
1:\(\sqrt{2}\):\(\sqrt{3}\)-1

Solution

The displacement x of a particle varies with time according to the relation x=\(\frac{a}{b}(1-e^{-bt})\) Then select the false alternatives.

At t=¹⁄_b, the displacement of the particle is nearly \(\frac{2}{3}\left ( \frac{a}{b} \right )\)
the velocity and acceleration ofthe particle at t = 0 area and –ab respectively
the particle cannot go beyond x=^a⁄_b
the particle will not come back to its starting point at t → ∞

Solution

The deceleration experienced by a moving motorboat after its engine is cut off, is given by dv/dt = – kv³ where k is constant. If v₀ is the magnitude of the velocity at cut-off,the magnitude of the velocity at a time t after the cut-off is

\(\frac{v_{0}}{\sqrt{(2v_{0}^{2}kt+1)}}\)
v₀e^-kt
v₀/2
v₀

Solution

The acceleration of aparticle is increasing lin early with time t as bt. The particle starts from the origin with an initial velocity v₀. The distance travelled by the particle in time t will be

V₀t+¹⁄₃bt²
V₀t+¹⁄₃bt³
V₀t+¹⁄₆bt³
V₀t+¹⁄₂bt²

Solution

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