Q. No.The displacement \(x\) of a particle varies with time \(t\) as \(x = ae^{-\alpha t}+ be^{\beta t}\), where \(a,\) \(b,\) \(\alpha,\) and \(\beta\) are positive constants. The velocity of the particle will:
1.
be independent of \(\alpha\) and \(\beta.\)
2.
go on increasing with time.
3.
drop to zero when \(\alpha=\beta.\)
4.
go on decreasing with time.
If a body travels some distance in a given time interval, then for that time interval, its:
| 1. | Average speed ≥ |Average velocity| |
| 2. | |Average velocity| ≥ Average speed |
| 3. | Average speed < |Average velocity| |
| 4. | |Average velocity| must be equal to average speed. |
When brakes are applied to a moving vehicle, the distance it travels before stopping is called stopping distance. It is an important factor for road safety and depends on the initial velocity \({v_0}\) and the braking capacity, or deceleration, \(-a\) that is caused by the braking. Expression for stopping distance of a vehicle in terms of \({v_0}\) and \(a\) is:
| 1. | \(\dfrac{{v_o}^2}{2a}\) | 2. | \(\dfrac{{v_o}}{2a}\) |
| 3. | \(\dfrac{{v_o}^2}{a}\) | 4. | \(\dfrac{2a}{{v_o}^2}\) |
The position of an object moving along the \(x\text-\)axis is given by, \(x=a+bt^2\), where \(a=8.5 ~\text m,\) \(b=2.5~\text{m/s}^2,\) and \(t\) is measured in seconds. Its velocity at \(t=2.0~\text s\) will be:
1. \(13~\text{m/s}\)
2. \(17~\text{m/s}\)
3. \(10~\text{m/s}\)
4. \(0~\text{m/s}\)
The figure shows the displacement-time graph of a particle moving on the x-axis. Then,
| 1. | the particle is continuously going in a positive x-direction. |
| 2. | the particle is at rest. |
| 3. | the velocity increases up to a time \(t_0\), and then becomes constant. |
| 4. | the particle moves at a constant velocity up to a time \(t_0\), and then stops. |
A stone is released from an elevator going up with an acceleration \(a.\) The acceleration of the stone after the release is:
1. \(a\) upward
2. \((g-a)\) upward
3. \((g-a)\) downward
4. \(g\) downward
Mark the correct statements for a particle going on a straight line:
| (a) | if the velocity and acceleration have opposite sign, the object is slowing down. |
| (b) | if the position and velocity have opposite sign, the particle is moving towards the origin. |
| (c) | if the velocity is zero at an instant, the acceleration should also be zero at that instant. |
| (d) | if the velocity is zero for a time interval, the acceleration is zero at any instant within the time interval. |
Choose the correct option:
| 1. | (a), (b) and (c) | 2. | (a), (b) and (d) |
| 3. | (b), (c) and (d) | 4. | all of these |
A boy throws a ball straight up the side of a building and receives it after \(4~\text s.\) On the other hand, if he throws it so that it strikes a ledge on its way up, it returns to him after \(3~\text s.\) The ledge is at a distance \(d\) below the highest point, where \(d=? \) (take acceleration due to gravity, \(g=10~\text{ms}^{-2})\)
1. \(5~\text m\)
2. \(2.5~\text m\)
3. \(1.25~\text m\)
4. \(10~\text m\)
1. statics
2. kinematics
3. mechanics
4. dynamics
When a particle's position changes, which of the following statements is true about its displacement and the distance it covers?
| 1. | Both cannot be zero. |
| 2. | Either one can be zero. |
| 3. | Both must be zero. |
| 4. | If one is positive, the other is negative, and vice-versa. |
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