A particle has an initial velocity (\(2\hat{i}+3\hat{j}\)) and an acceleration (\(0.3\hat{i}+0.2\hat{j}\)). 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. \(9~\text{units} \)

Subtopic:  Uniformly Accelerated Motion |
 88%
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A particle has an initial velocity \(\overrightarrow{u} = \left(4 \hat{i} - 5 \hat{j}\right)\) m/s and it is moving with an acceleration \(\overrightarrow{a} = \left(\frac{1}{4} \hat{i} + \frac{1}{5} \hat{j}\right)\text{m/s}^{2}\). Velocity of the particle at \(t=2\) s will be:
1. \((6\hat i -4\hat j)~\text{m/s}\)
2. \((4.5\hat i -4.5\hat j)~\text{m/s}\)
3. \((4.5\hat i -4.6\hat j)~\text{m/s}\)
4. \((6\hat i -4.6\hat j)~\text{m/s}\)

Subtopic:  Uniformly Accelerated Motion |
 86%
From NCERT
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A body started moving with an initial velocity of \(4\) m/s along the east and an acceleration \(1\) m/s2 along the north. The velocity of the body just after \(4\) s will be?

1. \(8\) m/s along East.
2. \(4 \sqrt{2} \) m/s along North-East.
3. \(8\) m/s along North.
4. \(4 \sqrt{2} \) m/s along South-East.
Subtopic:  Uniformly Accelerated Motion |
 78%
From NCERT
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A body sliding on a smooth inclined plane requires \(4\) seconds to reach the bottom starting from the rest at the top. How much time does it take to cover one-fourth distance starting from the rest at the top? 

1. \(1~\text{s}\) 2. \(2~\text{s}\)
3. \(4~\text{s}\) 4. \(16~\text{s}\)
Subtopic:  Uniformly Accelerated Motion |
 70%
From NCERT
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The time taken by a block of wood (initially at rest) to slide down a smooth inclined plane \(9.8~\text{m}\) long (angle of inclination is \(30^{\circ}\)) is:

               

1. \(\frac{1}{2}~\text{sec} \) 2. \(2 ~\text{sec} \)
3. \(4~ \text{sec} \) 4. \(1~\text{sec} \)
Subtopic:  Uniformly Accelerated Motion |
 70%
From NCERT
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A particle starts from the origin at \(t=0\) sec with a velocity of \(10\hat j~\text{m/s}\) and moves in the \(x\text-y\) plane with a constant acceleration of \((8.0\hat i +2.0 \hat j)~\text{m/s}^2\). At what time is the \(x\text-\)coordinate of the particle \(16\) m?
1. \(2\) s

2. \(3\) s

3. \(4\) s

4. \(1\) s

Subtopic:  Uniformly Accelerated Motion |
 70%
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A particle starts from the origin at \(t=0\) with a velocity of \(5.0\hat i\) m/s and moves in the \(x\text-y\) plane under the action of a force that produces a constant acceleration of \((3.0\hat i + 2.0\hat j)~\text{m/s}^2\). What is the speed of the particle at the instant its \(x\text-\)coordinate is \(84\) m?


1. \(36\) m/s
2. \(26\) m/s
3. \(1\) m/s
4. \(0\) m/s
Subtopic:  Uniformly Accelerated Motion |
 70%
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A particle starts moving with constant acceleration with initial velocity (\(\hat{\mathrm{i}}+5\hat{\mathrm{j}}\)) m/s. After \(4\) seconds, its velocity becomes (\(3\hat{\mathrm{i}}-2\hat{\mathrm{j}}\)) m/s. The magnitude of its displacement in 4 seconds is:
1. \(5\) m
2. \(10\) m
3. \(15\) m
4. \(20\) m

Subtopic:  Uniformly Accelerated Motion |
 65%
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A body is slipping from an inclined plane of height \(h\) and length \(l\). If the angle of inclination is \(\theta\), the time taken by the body to come from the top to the bottom of this inclined plane is:
1. \(\sqrt{\frac{2 h}{g}}\)
2. \(\sqrt{\frac{2 l}{g}}\)
3. \(\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)
4. \(\sin \theta \sqrt{\frac{2 h}{g}}\)

Subtopic:  Uniformly Accelerated Motion |
 59%
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A frictionless wire \(AB\) is fixed on a sphere of radius \(R\). A very small spherical ball slips on this wire. The time taken by this ball to slip from \(A\) to \(B\) is:
          
1. \(\frac{2 \sqrt{g R}}{g \cos \theta}\)
2. \(2 \sqrt{g R} . \frac{\cos \theta}{g}\)
3. \(2 \sqrt{\frac{R}{g}}\)
4. \(\frac{g R}{\sqrt{g\cos \theta}}\)

Subtopic:  Uniformly Accelerated Motion |
 53%
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