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In the following graph, the distance travelled by the body in metres is:

** **

1. | \(200\) | 2. | \(250\) |

3. | \(300\) | 4. | \(400\) |

Subtopic: Graphs |

86%

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Given below are two statements:

Assertion (A): |
Position-time graph of a stationary object is a straight line parallel to the time axis. |

Reason (R): |
For a stationary object, the position does not change with time. |

1. | Both (A) and (R) are True and (R) is the correct explanation of (A). |

2. | Both (A) and (R) are True but (R) is not the correct explanation of (A). |

3. | (A) is True but (R) is False. |

4. | Both (A) and (R) are False. |

Subtopic: Distance & Displacement |

88%

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The acceleration of a particle starting from rest varies with time according to the relation \(A= - a\omega^2\sin \omega t\). The displacement of this particle at a time *\(t\)* will be:

1. \(-\frac{1}{2}\left(a\omega^2\sin\omega t\right)t^2\)

2. \(a\omega \sin \omega t\)

3. \(a\omega \cos \omega t\)

4. \(a\sin \omega t\)

Subtopic: Non Uniform Acceleration |

60%

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A thief is running away on a straight road in a jeep moving with a speed of \(9\) m/s. A policeman chases him on a motorcycle moving at a speed of \(10\) m/s*.* If the instantaneous separation of the jeep from the motorcycle is \(100\) m, how long will it take for the policeman to catch the thief?

1. \(1\) s

2. \(19\) s

3. \(90\) s

4. \(100\) s

Subtopic: Relative Motion in One Dimension |

83%

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A particle is projected upwards. The times corresponding to height \(h\)* *while ascending and while descending are *t*_{1} and *t*_{2} respectively. The velocity of projection will be:

1. \(gt_1\)

2. \(gt_2\)

3. \(g(t_1+t_2)\)

4. \(\frac{g(t_1+t_2)}{2}\)

Subtopic: Uniformly Accelerated Motion |

69%

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The velocity-time \((v\text-t)\) graph of a body moving in a straight line is shown in the figure. The displacement and distance travelled by the body in \(6\) s* *are, respectively:

1. \(8\) m, \(16\) m

2. \(16\) m, \(8\) m

3. \(16\) m*,* \(16\) m

4. \(8\) m, \(8\) m

Subtopic: Graphs |

78%

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A stone dropped from a building of height \(h\) and reaches the earth after \(t\) seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity \(u\) and they reach the earth surface after \(t_1\) and \(t_2\) seconds respectively, then:

1. $t={t}_{1}-{t}_{2}$

2. $t=\frac{{t}_{1}+{t}_{2}}{2}$

3. $t=\sqrt{{t}_{1}{t}_{2}}$

4. $t={t}_{1}^{2}{t}_{2}^{2}$

Subtopic: Uniformly Accelerated Motion |

77%

From NCERT

PMT - 1997

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A particle moving in a straight line covers half the distance with a speed of \(3~\text{m/s}\). The other half of the distance is covered in two equal time intervals with speeds of \(4.5~\text{m/s}\)* *and \(7.5~\text{m/s}\)* *respectively. The average speed of the particle during this motion is:

1. \(4.0~\text{m/s}\)

2. \(5.0~\text{m/s}\)

3. \(5.5~\text{m/s}\)

4. \(4.8~\text{m/s}\)

Subtopic: Average Speed & Average Velocity |

70%

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A body starts to fall freely under gravity. The distances covered by it in the first, second and third second will be in the ratio:

1. \(1:3:5\)

2. \(1:2:3\)

3. \(1:4:9\)

4. \(1:5:6\)

Subtopic: Uniformly Accelerated Motion |

87%

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A student is standing at a distance of \(50\) metres from the bus. As soon as the bus begins its motion with an acceleration of \(1\) ms^{–2}, the student starts running towards the bus with a uniform velocity \(u\). Assuming the motion to be along a straight road, the minimum value of \(u\), so that the student is able to catch the bus is:

1. \(5\) ms^{–1}

2. \(8\) ms^{–1}

3. \(10\) ms^{–1}

4. \(12\) ms^{–1}

Subtopic: Uniformly Accelerated Motion |

73%

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