Differentiation
Trending Questions
Q.
The velocity of a particle is . Its position is at ; then its displacement after time is
Q. A particle moves along a straight line such that its displacement at any time t is given by s=(t3−6t2+3t+4) m. The velocity when the acceleration is zero is
- 3 m/s
- −12 m/s
- 42 m/s
- −9 m/s
Q.
The position of an elevator varies with time according to the function x(t)=t2+4t+3
What will be the velocity (in m/s) of the elevator at time t=4 s
Q.
If displacement of a moving particle in a plane at time is given by , then its acceleration is proportional to
Q. A particle moves along a straight line such that its displacement at any time t is given by s=(t3−6t2+3t+4) m. The velocity when the acceleration is zero is
- 3 m/s
- −12 m/s
- 42 m/s
- −9 m/s
Q. A particle moves in straight line such that its displacement x at time t is x2=t2+1. Acceleration of particle at t=34 sec is.
- 12564 m/s2
- 64125 m/s2
- 2516 m/s2
- 1625 m/s2
Q. The position of a particle is given by →r=3t^i+√3t2^j−4^k where t is in seconds and →r is meters. Find out magnitude of velocity →v and angle made by velocity →v with X-axis at t=√3 s.
- 3√5 m/s, θ=tan−1(2)
- 3√5 m/s, θ=tan−1(2√3)
- 3√2 m/s, θ=tan−1(3)
- 3√5 m/s, θ=tan−1(1√2)
Q. A body starts from orgin and moves along x-axis such that its velocity is v=(4t3−2t) m/s. Acceleration of particle when it is 2 m from origin is.
- 10 m/s2
- 20 m/s2
- 11 m/s2
- 22 m/s2
Q. A particle moves in straight line such that its displacement x at time t is x2=t2+1. Acceleration of particle at t=34 sec is.
- 12564 m/s2
- 64125 m/s2
- 2516 m/s2
- 1625 m/s2
Q. The position of a particle is given by →r=3t^i+√3t2^j−4^k, where t is in seconds and →r is in meters. Find out magnitude and direction of velocity →v with horizontal at t=√3 s.
- 3√5 m/s, θ=tan−1(2)
- 3√5 m/s, θ=tan−1(2√3)
- 3√2 m/s, θ=tan−1(3)
- 3√5 m/s, θ=tan−1(12)
Q. Power supplied to a particle of mass 2 kg varies with time as P=3t22 watt. Here t is in second. If velocity of particle at t=0 is v=0. The velocity of particle at time t=2 s will be
- 1 m/s
- 4 m/s
- 2 m/s
- 2√2 m/s
Q. The position of the particle is given by x(t)=(4t2−3t+2t3), its acceleration will be
- 4t+2t2
- 12t2+8t
- 8t−3+6t2
- 12t+8
Q. A particle moves along parabolic path x=y2+2y+2 in such a way that y− component of velocity is constant and equal to 5 m/s during the motion. Magnitude of acceleration of the particle
- 50 m/s2
- 100 m/s2
- 10√2 m/s2
- 10 m/s2
Q. A particle travels according to the equation y=x−x22. Find the maximum height it achieves.
- 1 m
- 2 m
- 12 m
- 14 m
Q. A function is given as y=5x2−10x, then
- Maximum value of y is 10
- Minimum value of y is -5
- Maximum value of y is 0
- Maximum value of y occurs at x = 1
Q. The position of a particle is given by →r=3t^i+√3t2^j−4^k, where t is in seconds and →r is in meters. Find out magnitude and direction of velocity →v with horizontal at t=√3 s.
- 3√5 m/s, θ=tan−1(2)
- 3√5 m/s, θ=tan−1(2√3)
- 3√2 m/s, θ=tan−1(3)
- 3√5 m/s, θ=tan−1(12)
Q. The position vector of a moving particle in x−y plane can be represented as →r=(2tˆi+2t2ˆj)m, where t represents the time instant. The rate of change of θ at time t=2s. (where θ is the angle which its velocity vector makes with positive x−axis)
- 25 rad/s
- 217 rad/s
- 2 rad/s
- 4 rad/s