Q. No.Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time \(t_1.\) On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time \(t_2.\) The time taken by her to walk upon the moving escalator will be:
1.
\(\dfrac{t_1t_2}{t_2-t_1}\)
2.
\(\dfrac{t_1t_2}{t_2+t_1}\)
3.
\(t_1-t_2\)
4.
\(\dfrac{t_1+t_2}{2}\)
If the velocity of a particle is \(v=At+Bt^{2},\) where \(A\) and \(B\) are constants, then the distance travelled by it between \(1~\text{s}\) and \(2~\text{s}\) is:
| 1. | \(3A+7B\) | 2. | \(\frac{3}{2}A+\frac{7}{3}B\) |
| 3. | \(\frac{A}{2}+\frac{B}{3}\) | 4. | \(\frac{3A}{2}+4B\) |
1. \(- 2 nβ^{2} x^{- 2 n - 1}\)
2. \(- 2 nβ^{2} x^{- 4 n - 1}\)
3. \(- 2 \beta^{2} x^{- 2 n + 1}\)
4. \(- 2 nβ^{2} x^{- 4 n + 1}\)
A particle is moving such that its position coordinates (x, y) are (\(2\) m, \(3\) m) at time \(t=0,\) (\(6\) m,\(7\) m) at time \(t=2\) s, and (\(13\) m, \(14\) m) at time \(t=\) \(5\) s. The average velocity vector \(\vec{v}_{avg}\) from \(t=\) 0 to \(t=\) \(5\) s is:
1. \({1 \over 5} (13 \hat{i} + 14 \hat{j})\)
2. \({7 \over 3} (\hat{i} + \hat{j})\)
3. \(2 (\hat{i} + \hat{j})\)
4. \({11 \over 5} (\hat{i} + \hat{j})\)
A stone falls freely under gravity. It covers distances \(h_1,~h_2\) and \(h_3\) in the first \(5\) seconds, the next \(5\) seconds and the next \(5\) seconds respectively. The relation between \(h_1,~h_2\) and \(h_3\) is:
| 1. | \(h_1=\frac{h_2}{3}=\frac{h_3}{5}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \) |
| 2. | \(h_2=3h_1\) and \(h_3=3h_2\) |
| 3. | \(h_1=h_2=h_3\) |
| 4. | \(h_1=2h_2=3h_3\) |
A particle has initial velocity \(\left(2 \hat{i} + 3 \hat{j}\right)\) and acceleration \(\left(0 . 3 \hat{i} + 0 . 2 \hat{j}\right)\). The magnitude of velocity after \(10\) s will be:
| 1. | \(9 \sqrt{2}~ \text{units}\) | 2. | \(5 \sqrt{2} ~\text{ units}\) |
| 3. | \(5~\text{units}\) | 4. | \(9~\text{units}\) |
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
| 1. | 20 m/s | 2. | 40 m/s |
| 3. | 5 m/s | 4. | 10 m/s |
A ball is dropped from a high-rise platform at \(t=0\) starting from rest. After \(6\) seconds, another ball is thrown downwards from the same platform with speed \(v\). The two balls meet after \(18\) seconds. What is the value of \(v\)?
| 1. | \(75\) ms-1 | 2. | \(55\) ms-1 |
| 3. | \(40\) ms-1 | 4. | \(60\) ms-1 |
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. (velocity)\(3/2\)
2. (distance)\(2\)
3. (distance)\(-2\)
4. (velocity)\(2/3\)
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