NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI03:MOTION IN A STRAIGHT LINE
362171
The velocity of the particle at any time \(t\) is given by \(V=2 t(3-t) m s^{-1}\). At what time is its velocity maximum?
1 2 second
2 1.5 second
3 1 second
4 5 second
Explanation:
At the time of maximum velocity, acceleration is zero. \(V=2 t(3-t) \quad V=6 t-2 t^{2}\) Differentiating both side \(\dfrac{d V}{d t}=\dfrac{d\left(6 t-2 t^{2}\right)}{d t} \quad o=6-4 t \quad t=1.5\)
PHXI03:MOTION IN A STRAIGHT LINE
362172
Assertion : In a uniformly accelerated motion,acceleration time graph is a straight line with positive slope. Reason : Acceleration is rate of change of velocity
1 Both assertion and reason are correct and reason is the correct explanation of assertion
2 Both assertion and reason are correct but reason is not the correct explanation of assertion
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Positive slope indicates that acceleration increases uniformly with time. It is not uniform.So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362174
The position \(x\) of a particle varies with time (\(t\)) as \(x = A{t^2} - B{t^3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\)?
1 \(\frac{{2\;A}}{{3\;B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{A}{{3\;B}}\)
4 zero
Explanation:
Given that \(x = A{t^2} - B{t^3}\) \(\therefore\) velocity \( = \frac{{dx}}{{dt}} = 2At - 3B{t^2}\) and acceleration \( = \frac{d}{{dt}}\left( {\frac{{dx}}{{dt}}} \right) = 2\;A - 6Bt\) For acceleration to be zero \(2\;A - 6Bt = 0\). \(\therefore t = \frac{{2\;A}}{{6\;B}} = \frac{A}{{3\;B}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362175
The velocity - time graph for two bodies \(A\) and \(B\) are shown. Then the acceleration of \(A\) and \(B\) are in the ratio
1 \(\tan 25^\circ \,{\rm{to}}\,\tan 50^\circ \)
2 \(\cos 25^\circ \,{\rm{to}}\,\cos 50^\circ \)
3 \(\tan 25^\circ \,{\rm{to}}\,\tan 40^\circ \)
4 \(\sin 25^\circ \,{\rm{to}}\,\sin 40^\circ \)
Explanation:
As the slope of \(v - t\) graph gives acceleration, so acceleration \(a = \tan \theta \) \(\therefore \) From the given graph The acceleration of \(A\) is \({a_A} = \tan 25^\circ \) and that of \(B\) is \({a_B} = \tan \,(90^\circ - 40^\circ ) = \tan 50^\circ \) \(\frac{{{a_A}}}{{{a_B}}} = \frac{{\tan 25^\circ }}{{\tan 50^\circ }}\)
362171
The velocity of the particle at any time \(t\) is given by \(V=2 t(3-t) m s^{-1}\). At what time is its velocity maximum?
1 2 second
2 1.5 second
3 1 second
4 5 second
Explanation:
At the time of maximum velocity, acceleration is zero. \(V=2 t(3-t) \quad V=6 t-2 t^{2}\) Differentiating both side \(\dfrac{d V}{d t}=\dfrac{d\left(6 t-2 t^{2}\right)}{d t} \quad o=6-4 t \quad t=1.5\)
PHXI03:MOTION IN A STRAIGHT LINE
362172
Assertion : In a uniformly accelerated motion,acceleration time graph is a straight line with positive slope. Reason : Acceleration is rate of change of velocity
1 Both assertion and reason are correct and reason is the correct explanation of assertion
2 Both assertion and reason are correct but reason is not the correct explanation of assertion
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Positive slope indicates that acceleration increases uniformly with time. It is not uniform.So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362174
The position \(x\) of a particle varies with time (\(t\)) as \(x = A{t^2} - B{t^3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\)?
1 \(\frac{{2\;A}}{{3\;B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{A}{{3\;B}}\)
4 zero
Explanation:
Given that \(x = A{t^2} - B{t^3}\) \(\therefore\) velocity \( = \frac{{dx}}{{dt}} = 2At - 3B{t^2}\) and acceleration \( = \frac{d}{{dt}}\left( {\frac{{dx}}{{dt}}} \right) = 2\;A - 6Bt\) For acceleration to be zero \(2\;A - 6Bt = 0\). \(\therefore t = \frac{{2\;A}}{{6\;B}} = \frac{A}{{3\;B}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362175
The velocity - time graph for two bodies \(A\) and \(B\) are shown. Then the acceleration of \(A\) and \(B\) are in the ratio
1 \(\tan 25^\circ \,{\rm{to}}\,\tan 50^\circ \)
2 \(\cos 25^\circ \,{\rm{to}}\,\cos 50^\circ \)
3 \(\tan 25^\circ \,{\rm{to}}\,\tan 40^\circ \)
4 \(\sin 25^\circ \,{\rm{to}}\,\sin 40^\circ \)
Explanation:
As the slope of \(v - t\) graph gives acceleration, so acceleration \(a = \tan \theta \) \(\therefore \) From the given graph The acceleration of \(A\) is \({a_A} = \tan 25^\circ \) and that of \(B\) is \({a_B} = \tan \,(90^\circ - 40^\circ ) = \tan 50^\circ \) \(\frac{{{a_A}}}{{{a_B}}} = \frac{{\tan 25^\circ }}{{\tan 50^\circ }}\)
362171
The velocity of the particle at any time \(t\) is given by \(V=2 t(3-t) m s^{-1}\). At what time is its velocity maximum?
1 2 second
2 1.5 second
3 1 second
4 5 second
Explanation:
At the time of maximum velocity, acceleration is zero. \(V=2 t(3-t) \quad V=6 t-2 t^{2}\) Differentiating both side \(\dfrac{d V}{d t}=\dfrac{d\left(6 t-2 t^{2}\right)}{d t} \quad o=6-4 t \quad t=1.5\)
PHXI03:MOTION IN A STRAIGHT LINE
362172
Assertion : In a uniformly accelerated motion,acceleration time graph is a straight line with positive slope. Reason : Acceleration is rate of change of velocity
1 Both assertion and reason are correct and reason is the correct explanation of assertion
2 Both assertion and reason are correct but reason is not the correct explanation of assertion
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Positive slope indicates that acceleration increases uniformly with time. It is not uniform.So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362174
The position \(x\) of a particle varies with time (\(t\)) as \(x = A{t^2} - B{t^3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\)?
1 \(\frac{{2\;A}}{{3\;B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{A}{{3\;B}}\)
4 zero
Explanation:
Given that \(x = A{t^2} - B{t^3}\) \(\therefore\) velocity \( = \frac{{dx}}{{dt}} = 2At - 3B{t^2}\) and acceleration \( = \frac{d}{{dt}}\left( {\frac{{dx}}{{dt}}} \right) = 2\;A - 6Bt\) For acceleration to be zero \(2\;A - 6Bt = 0\). \(\therefore t = \frac{{2\;A}}{{6\;B}} = \frac{A}{{3\;B}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362175
The velocity - time graph for two bodies \(A\) and \(B\) are shown. Then the acceleration of \(A\) and \(B\) are in the ratio
1 \(\tan 25^\circ \,{\rm{to}}\,\tan 50^\circ \)
2 \(\cos 25^\circ \,{\rm{to}}\,\cos 50^\circ \)
3 \(\tan 25^\circ \,{\rm{to}}\,\tan 40^\circ \)
4 \(\sin 25^\circ \,{\rm{to}}\,\sin 40^\circ \)
Explanation:
As the slope of \(v - t\) graph gives acceleration, so acceleration \(a = \tan \theta \) \(\therefore \) From the given graph The acceleration of \(A\) is \({a_A} = \tan 25^\circ \) and that of \(B\) is \({a_B} = \tan \,(90^\circ - 40^\circ ) = \tan 50^\circ \) \(\frac{{{a_A}}}{{{a_B}}} = \frac{{\tan 25^\circ }}{{\tan 50^\circ }}\)
362171
The velocity of the particle at any time \(t\) is given by \(V=2 t(3-t) m s^{-1}\). At what time is its velocity maximum?
1 2 second
2 1.5 second
3 1 second
4 5 second
Explanation:
At the time of maximum velocity, acceleration is zero. \(V=2 t(3-t) \quad V=6 t-2 t^{2}\) Differentiating both side \(\dfrac{d V}{d t}=\dfrac{d\left(6 t-2 t^{2}\right)}{d t} \quad o=6-4 t \quad t=1.5\)
PHXI03:MOTION IN A STRAIGHT LINE
362172
Assertion : In a uniformly accelerated motion,acceleration time graph is a straight line with positive slope. Reason : Acceleration is rate of change of velocity
1 Both assertion and reason are correct and reason is the correct explanation of assertion
2 Both assertion and reason are correct but reason is not the correct explanation of assertion
3 Assertion is correct but reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Positive slope indicates that acceleration increases uniformly with time. It is not uniform.So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362174
The position \(x\) of a particle varies with time (\(t\)) as \(x = A{t^2} - B{t^3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\)?
1 \(\frac{{2\;A}}{{3\;B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{A}{{3\;B}}\)
4 zero
Explanation:
Given that \(x = A{t^2} - B{t^3}\) \(\therefore\) velocity \( = \frac{{dx}}{{dt}} = 2At - 3B{t^2}\) and acceleration \( = \frac{d}{{dt}}\left( {\frac{{dx}}{{dt}}} \right) = 2\;A - 6Bt\) For acceleration to be zero \(2\;A - 6Bt = 0\). \(\therefore t = \frac{{2\;A}}{{6\;B}} = \frac{A}{{3\;B}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362175
The velocity - time graph for two bodies \(A\) and \(B\) are shown. Then the acceleration of \(A\) and \(B\) are in the ratio
1 \(\tan 25^\circ \,{\rm{to}}\,\tan 50^\circ \)
2 \(\cos 25^\circ \,{\rm{to}}\,\cos 50^\circ \)
3 \(\tan 25^\circ \,{\rm{to}}\,\tan 40^\circ \)
4 \(\sin 25^\circ \,{\rm{to}}\,\sin 40^\circ \)
Explanation:
As the slope of \(v - t\) graph gives acceleration, so acceleration \(a = \tan \theta \) \(\therefore \) From the given graph The acceleration of \(A\) is \({a_A} = \tan 25^\circ \) and that of \(B\) is \({a_B} = \tan \,(90^\circ - 40^\circ ) = \tan 50^\circ \) \(\frac{{{a_A}}}{{{a_B}}} = \frac{{\tan 25^\circ }}{{\tan 50^\circ }}\)
