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Q. No.A particle moves a distance \(x\) in time \(t\) according to equation \(x = (t+5)^{-1}\). The acceleration of the particle is proportional to:
| 1. | \((\text{velocity})^{\frac{3}{2}}\) | 2. | \((\text{distance})^2\) |
| 3. | \((\text{distance})^{-2}\) | 4. | \((\text{velocity})^{\frac{2}{3}}\) |
When the velocity of a body is variable, then:
| 1. | its speed may be constant |
| 2. | its acceleration may be constant |
| 3. | its average acceleration may be constant |
| 4. | all of the above |
A body is projected vertically in the upward direction from the surface of the earth. If the upward direction is taken as positive, then the acceleration of the body during its upward and downward journey is:
| 1. | Positive, negative | 2. | Negative, negative |
| 3. | Positive, positive | 4. | Negative, positive |
The velocity \(v\) of an object varies with its position \(x\) on a straight line as \(v=3\sqrt{x}.\) Its acceleration versus time \((a\text-t)\) graph is best represented by:
| 1. |  |
2. |  |
| 3. |  |
4. |  |
1. \(\dfrac{5}{3}\) m/s
2. \(\dfrac{4}{3}\) m/s
3. \(2\) m/s
4. \(3\) m/s
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 Cheetah can accelerate from \(0\) to \(96\) km/h in \(2\) s. What is the average acceleration of the Cheetah?
1. \(10\) m/s2
2. \(13.3\) m/s2
3. \(15\) m/s2
4. \(48\) m/s2
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 motion of a particle along a straight line is described by the equation \(x = 8+12t-t^3\) where \(x \) is in meter and \(t\) in seconds. The retardation of the particle, when its velocity becomes zero, is:
1. \(24\) ms-2
2. zero
3. \(6\) ms-2
4. \(12\) ms-2
Then its acceleration varies as:
1. \(\dfrac{1}{\sqrt x}\)
2. \(x^{3/2}\)
3. \(x^{-3/2}\)
4. \(x^0\)
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