Motion Under Variable Acceleration
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- Direction of motion is not changing at any of the instants
- Direction of motion is changing at t=3 s
- For 0<t<3 s, the particle is slowing down
- For 0<t<3 s the particle is speeding up
- 54 m
- 81 m
- 24 m
- 32 m
A particle retards from a velocity v0 while moving in a straight line. If the magnitude of deceleration is directly proportional to the square root of the speed of the particle, find its average velocity for the total time of its motion.
2v03
−2v03
v03
−v03
- 2a3b
- ab
- a3b
- Zero
- 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
- Increases linearly with x
- Constant
- Increases parabolically with x
- None
A particle moves according to the equation dvdt = α - β v , where α and β are constants. Find the velocity as a funtion of time. Assume body starts from rest.
v = (βα) (1 - e−βt)
v = (βα) (e−βt)
v = (αβ) (1 - e−βt)
v = (αβ) (e−βt)
A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s2. He reaches the ground with a speed of 3 m/s. At what height, did he bail out?
293 m
111 m
91 m
182 m
The displacement x of a particle along a straight line at time t is given by x=a0+a1t+a2t2.The acceleration of the particle is
a0
a1
2a2
a2
- The acceleration of the particle is zero at t = 0 second
- The velocity of the particle is zero at t = 0 second
- The velocity of the particle is zero at t = 1 second
- The velocity and acceleration of the particle are never zero
Referring to a - x graph, find the velocity when the displacement of the particle is 100 m. assume initial velocity as zero.
9.8 m/s
-10 √6 m/s
20 m/s
10 √6 m/s
- 16 m
- 64 m
- 48 m
- 32 m
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 to t = 1 s
t = 0.5 s to t = 1 s
t = 0 to t = 0.5 s
t > 1 s
The position vector and acceleration vector are perpendicular
- at t=1 s
- at t=0
- at t=√2 s
- at t=1.5 s
The position vector and acceleration vector are parallel
- at t=1 s
- at t=0
- at t=√2 s
- at t=1.5 s
The position vector and acceleration vector are parallel
- at t=1 s
- at t=0
- at t=√2 s
- at t=1.5 s
- kt2+ct
- 12(kt2+ct)
- 12(kt2+4ct)
- kt22+ct
- Particle will cross the origin at t=2√2 s.
- Magnitude of velocity and acceleration are equal at t=1 s
- Particle will cross the origin at t=5√2 s.
- Magnitude of velocity and acceleration are equal at t=3 s.
- 2a3b
- ab
- a3b
- Zero
- 325
- 2425
- 18
- 328
- kt2+ct
- 12(kt2+ct)
- 12(kt2+4ct)
- kt22+ct
The velocity of a particle moving along a straight line increases according to the linear law v = v0 + kx
where k is a constant . Then
1) the accelearation of the particle is k(v0 + kx)
2) the particle takes a time 1k loge (v1v0) to attain a velocity v1
3) velocity varies linearly with displacement with slope of velocity displacement curve equal to k
4) data is insufficient to arrive at a conclusion.
1, 2, 3 are correct
2, 3 are correct
all of them are correct
1 and 2 are correct
- 28 m/s2
- 22 m/s2
- 12 m/s2
- 10 m/s2
- The initial velocity of particle is 4
- The acceleration of particle is 2a.
- The particle is at origin at t = 0
- None of these
An ant starts from the origin and walks along a straight line. After every second, its position (in mm) is observed by a curious 5-year old. These are his readings:
Position (mm) 0 10 40 100 200
After time (s) 0 1 2 3 4
By looking at this table, which of the following could be a possible description of the ant's motion is
(a) Uniform velocity
(b) uniform acceleration (with non-uniform velocity)
(c) Uniform jerk (with non-uniform accelerator)
(d) cannot be determined (can be any of the above)
Uniform velocity
uniform acceleration (with non-uniform velocity
Uniform jerk (with non-uniform accelerator)
cannot be determined (can be any of the above)
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
A body is moving in a straight line, starting its journey from origin. At any instant, its velocity is given by, K1t(K1−1) when K1 is a constant and t is the time. Find the acceleration of the particle, when it is at a distance s from the origin:
K1(K1−1)s(1−1/K1)
K1(K1−1)s(1−2/K1)
K1(K1−1)s(2−1/K1)
K1(K1−1)s(3−1/K1)
If the speedometer(in km/h) of a bike riding along a straight road varies with time as given V(t)=3t2
What will be the distance it has covered after time t = 2 hr?