Motion Under Variable Acceleration
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The displacement x of a particle moving in one dimension is related to time t by the equation t=√x+3 where x is in metres and t in seconds. The displacement of the particle when its velocity is zero is
zero
4 m
1 m
0.5 m
A particle moves in a straight line with a constant acceleration. It changes its velocity from 10 ms-1 to 20 ms-1 while passing through a distance of 135 m in âtâ seconds. The value of t is:
12
9
10
1.8
- (velocity)3/2
- (distance)2
- (velocity)2/3
- (distance)−2
- 54 m
- 81 m
- 24 m
- 32 m
- 2βv3
- 2αβv3
- 2β2v3
- 2αv3
A particle has initial velocity and acceleration the magnitude of the velocity after will be:
An object moving with a speed of 6.25 m/s, is decelerated at a rate given by:
dvdt=−2.5√v
where v is instantaneous speed. The time taken by the object, to come to rest, would be:
- 2 s
- 4 s
- 1 s
- 8 s
The position of a particle varies with time as . The acceleration of a particle is zero at time equal to
- t−1/2
- t1/2
- t/√m
- t2P0
- v0t+13bt2
- v0t+13bt3
- v0t+16bt3
- v0t+12bt2
The velocity of a body depends on time according to equation V=20 + 0.1 t2. the body is undergoing
Uniform Acceleration
Uniform Retardation
Non-Uniform acceleration
Zero Acceleration
The speed-time graph of a particle moving along a fixed direction is shown in Fig. 3.28.
Obtain the distance traversed by the particle between (a) t = 0 s to 10 s, (b) t = 2 s to 6 s.
What is the average speed of the particle over the intervals in (a) and (b)?
The speed verses time graph for a particle is shown in the figure. The distance travelled (in m) by the particle during the time interval will be.
The displacement of a particle varies with time t, x=ae−at+beβ t where a, b, α and β are positive constants. The velocity of the particle will?
The acceleration of a particle, starting from rest, varies with time according to the relation a = Kt + C, where K and C are constants of motion. The velocity ‘V’ of the particle after a time ‘t’ will be
- Kt+Ct
- 12Kt2+Ct
- 12(Kt2+Ct)
- Kt2+12Ct
- (a+c)
- 2b
- 6a
- (a+b)
The equation of motion of a particle is given by , where s and t are measured in cm and sec. The time when the particle stops momentarily is
None of the above
The position vector and acceleration vector are perpendicular
- at t=√2 s
- at t=1.5 s
- at t=1 s
- at t=0
A particle starts moving rectilinearly at time t = 0 such that its velocity ‘v’ changes with time ‘t’ according to the equation v=t2−t where t is in seconds and v in m/s. Find the time interval for which the particle retards.
t = 0.5 s to t = 1 s
t = 0 to t = 0.5 s
t > 1 s
t = 0 to t = 1 s
A particle is moving along the x-axis with its coordinate with the time given by meter. Another particle is moving along the y-axis with its coordinate as a function of time given by meter. . At second , the speed of the second particle as measured in the frame of the first particle is given as . Then is
The displacement x of a particle varies with time according to the relation x=ab(1−e−bt). Which of the following statements is correct?
At t=1b, the displacement of the particle is ab.
The velocity and acceleration of the particle at t = 0 are a and b respectively.
The particle will come back to its starting point as t→∞.
None of these
- The particle will cover a total distance v0α
- The particle will come to rest after a time 1α
- The particle will continue to move for a very long time
- The velocity of the particle will become v02 after a time 1α
- 65 m/s
- 55 m/s
- 45 m/s
- 35 m/s
- 1aln(1+aut)
- aln(aut)
- aut
- 1aln(aut)
- −9x m/s2
- −18x m/s2
- −9x2 m/s2
- None of these
- 2100msec2 downwards
- 1400msec2
- 2100msec2 downwards
- 700msec2
- Average velocity is 253 m/s
- Average speed is 10 m/s
- Average velocity is 53 m/s
- Acceleration is 4 m/s2 at t = 0
- 6a
- 2b
- (a+b)
- (a+c)