Instantaneous acceleration
Trending Questions
Q. The velocity of the particle is given by v=3t2+2t in m/s. The acceleration and displacement of the particle as a function of time respectively are
[Take x=0 at t=0]
[Take x=0 at t=0]
- (t+2) m/s2, (3t3+t2) m
- (6t+2) m/s2, (t3+t2) m
- (6t2+2) m/s2, (t3+t2) m
- (t+2) m/s2, (t3+t) m
Q. A particle is moving along a straight-line path according to the relation
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
- S−3
- S3/2
- S−2/3
- S2
Q. A particle moves along X− axis. It's equation of motion is x=2t(t−2) where t is in seconds. Choose the correct statement(s).
- The particle moves with a non-uniform acceleration.
- The particle momentarily comes to rest at t=1 sec.
- The particle performs SHM
- The given equation represents uniformly accelerated motion for t>1 sec
Q. A particle moves along X− axis. It's equation of motion is x=2t(t−2) where t is in seconds. Choose the correct statement(s).
- The particle moves with a non-uniform acceleration.
- The particle momentarily comes to rest at t=1 sec.
- The particle performs SHM
- The given equation represents uniformly accelerated motion for t>1 sec
Q. A car accelerates uniformly from 13 ms–1 to 31 ms–1 while entering the motorway, covering a distance of 220 m. Then the acceleration of the car will be:
- 2.9 ms−2
- 1.8 ms−2
- 4 ms−2
- 2.2 ms−2
Q. The motion of a body is given by the equation dv(t)dt=6.0−3v(t). Where v(t) is speed in ms and t in sec. If body was at rest at t = 0
- The terminal speed is 3.0ms
- The speed varies with the time as v(t)=2(1−e−3t)ms
- The speed is 0.1ms when the acceleration is half the initial value
- The magnitude of the initial acceleration is 0.5ms2
Q. Velocity (V) versus displacement (S) graph of a particle moving in a straight line is as shown in the figure. Given Vo=4 m/s and So=20 m.
Corresponding acceleration (a) versus displacement (S) graph of the particle would be
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/700232/original_7.que.PNG)
Corresponding acceleration (a) versus displacement (S) graph of the particle would be
Q. Velocity (V) versus displacement (S) graph of a particle moving in a straight line is as shown in the figure. Given V0=4 m/s and So=20 m. At S=30 m, what is the acceleration of particle?
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/700240/original_7.que.PNG)
- 0.2 m/s2
- 0.4 m/s2
- −0.2 m/s2
- −0.4 m/s2
Q. A Porsche undergoing uniformly accelerated motion can go from 0 km/h to 100 km/h in 10 s. What is its instantaneous acceleration?
- 2.77 m/s2
- 3.66 m/s2
- −3.66 m/s2
- 0 m/s2
Q. A particle is moving along a straight-line path according to the relation
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
- S−3
- S3/2
- S−2/3
- S2
Q. The displacement travelled by a particle in a straight-line motion is directly proportional to t1/2, where t= time elapsed. What is the nature of motion?
- Increasing acceleration
- Decreasing acceleration
- Increasing retardation
- Decreasing retardation
Q. Velocity of a car as a function of time is given by v(t)=(t3−cost) in m/s. The acceleration in (m/s2) of the car as a function of time t is
- 3t2+cost
- t22+sint
- 3t2+sint
- t22−cost
Q. A point moves in a straight line so that its displacement x is given by x2=1+t2 where, x is in m and time t is in s. Its acceleration in m/s2 at time t is
- tx3
- 1x3
- x−1x3
- t+1x3
Q. A body initially moving with a velocity of 5 ms–1, attains a velocity of 25 ms–1 in 5 s. The acceleration of the body is:
(Assume constant acceleration)
(Assume constant acceleration)
- 8 ms−2
- 7 ms−2
- 4 ms−2
- 3 ms−2
Q. The displacement is given by x=2t2+t+5 in m, the acceleration at t=5s will be
- 8 m/s2
- 12 m/s2
- 15 m/s2
- 4 m/s2
Q. The acceleration of a particle is increasing linearly with time t as bt . The particle starts from the origin with an initial velocity vo . The distance travelled by the particle in time t will be
- v0t+13bt2
- v0t+13bt3
- v0t+16bt3
- v0t+12bt2
Q.
A body starts from the origin and moves along the X-axis such that the velocity at any instant is given by (4t3−2t), where t is in sec and velocity in m/s. What is the acceleration of the particle, when it is 2 m from the origin
28m/s2
22m/s2
12m/s2
10m/s2
Q. 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
- Go on decreasing with time
- Be independent of α and β .
- Drop to zero when α=β
- Go on increasing with time
Q. A particle is moving along a straight-line path according to the relation
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
S2=at2+2bt+c
S represents the displacement covered in t seconds and a, b, c are constants. The acceleration of the particle varies as
- S−2/3
- S3/2
- S2
- S−3
Q. The velocity of a particle moving in the positive direction of x− axis varies as v=10√x (v is in ms−1, x is in m). Assuming that at t=0, particle was at x=0. Then,
- The initial velocity of the particle is zero
- The initial velocity of the particle is 2.5 ms−1
- The acceleration of the particle is 2.5 ms−2
- The acceleration of the particle is 50 ms−2
Q. A particle is moving along x-axis whose instantaneous speed is v=√108−9x2. The acceleration of the particle is
- −9x m/s2
- −18x m/s2
- −9x2 m/s2
- None of these
Q. 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
Q. The velocity- time graph of a body is given in figure. The maximum acceleration in ms−2 is
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/693743/original_15.png)
![](https://s3-us-west-2.amazonaws.com/infinitestudent-images/ckeditor_assets/pictures/693743/original_15.png)
- 4
- 3
- 2
- 1
Q. The speed v of a car moving on a straight road changes according to equation, v2=a+bx, where a and b are positive constants. Then, the magnitude of acceleration in the course of such motion: (x is the distance travelled) :
- Increases
- Decreases
- Stays constant
- First decreases and then increases
Q. The relation between time t and distance x is t=αx2+βx where α and β are constants. The retardation is
- 2 αv3
- 2 βv3
- 2 αβv3
- 2 β2v3
Q. The graph shows the variation of 1v (where v is the velocity of the particle) with respect to time. Then find the value of acceleration at t=3 sec
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1046669/original_Untitled-2.png)
![](https://df0b18phdhzpx.cloudfront.net/ckeditor_assets/pictures/1046669/original_Untitled-2.png)
- 3 m/s2
- 5 m/s2
- 1 m/s2
- 0.5 m/s2
Q. 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
- Go on decreasing with time
- Be independent of α and β .
- Drop to zero when α=β
- Go on increasing with time
Q. A particle is moving along a curve . Then
- if its speed is constant, it has no acceleration
- if its speed is increasing, the acceleration of the particle is along its direction of motion
- The direction of its acceleration cannot be along the tangent.
- None of these
Q. The position x of a particle varies with time t as x=αt2−βt3. Which of the following is (are) correct?
- The particle will return to its starting point after time αβ.
- The particle will come to rest after time 2α3β.
- The initial velocity of the particle was zero but its initial acceleration was not zero.
- The acceleration of the particle is zero at t=α3β.