NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI03:MOTION IN A STRAIGHT LINE
362248
The position of a particle along \(x\)-axis at time \(t\) is given by \(x = 2 + t - 3{t^2}\). The displacement and the distance travelled in the interval \(t = 0\,{\rm{and}}\,t = 1\) are respectively
1 \(2,2\)
2 \( - 2,2.5\)
3 \(0,2\)
4 \( - 2,2.1\)
Explanation:
\(x = 2 + t - 3{t^2},\;v = \frac{{dx}}{{dt}} = 1 - 6t,\) velocity will become zero at time, \(0 = 1 - 6 {t_0}\;{\rm{or}}\;{t_0} = \frac{1}{6}s,\) Since the given time \(t = 1\,s\) is greater than \({t_0} = \frac{1}{6}s,\) \({\rm{distance}} > \left| {{\rm{displacement}}} \right|\) Displacement \(S = {x_f} - {x_i}\) \( = (2 + 1 - 3) - (2 + 0 - 0) = - 2\;m\) Distance \(d = \left| {{S_0} - {t_0}} \right| + \left| {{S_t}_{ - {t_0}}} \right|\) \( = \frac{{{u^2}}}{{2\left| a \right|}} + \frac{1}{2}\left| a \right|{\left( {t - {t_0}} \right)^2}\) Comparing \(v = 1 - 6t\) with \(v = u + ut,\) we have \(u = 1\;m{\rm{/}}s\,{\rm{and}}\,a = - 6\;m{\rm{/}}{s^2}\) \(\therefore {\rm{Distance}} = \frac{{{{\left( 1 \right)}^2}}}{{2 \times 6}} + \frac{1}{2} \times 6 \times {\left( {1 - \frac{1}{6}} \right)^2}\) \( = 2.1\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362249
Assertion : The slope of displacement-time of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason : Slope of displacement-time graph = Velocity of the body.
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:
Slope of displacement-time measures velocity of an object. So option (1) is correct
PHXI03:MOTION IN A STRAIGHT LINE
362250
A wheel of radius 1 \(m\) rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
1 \(2\pi \)
2 \(\sqrt 2 \pi \)
3 \(\sqrt {{\pi ^2} + 4} \)
4 \(\pi \)
Explanation:
Horizontal distance by the wheel in half revolution \( = \pi R\) So the displacement of the point which was initially in contact with ground \( = AA' = \sqrt {{{\left( {\pi R} \right)}^2} + {{\left( {2R} \right)}^2}} \) \( = R\sqrt {{\pi ^2} + 4} \) \( = \sqrt {{\pi ^2} + 4} \; ({\rm{as}} \, R = 1 \, m)\)
PHXI03:MOTION IN A STRAIGHT LINE
362251
A person moves towards east for \(4 \, m\), then towards north for \(3 \, m\) and then moves vertically up by \(5 \, m\). What is his distance now from the starting point?
1 \(5\sqrt 2 \, m\)
2 \(5 \, m\)
3 \(10 \, m\)
4 \(20 \, m\)
Explanation:
Distance from starting point \(\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = 5\sqrt 2 \,m\)
362248
The position of a particle along \(x\)-axis at time \(t\) is given by \(x = 2 + t - 3{t^2}\). The displacement and the distance travelled in the interval \(t = 0\,{\rm{and}}\,t = 1\) are respectively
1 \(2,2\)
2 \( - 2,2.5\)
3 \(0,2\)
4 \( - 2,2.1\)
Explanation:
\(x = 2 + t - 3{t^2},\;v = \frac{{dx}}{{dt}} = 1 - 6t,\) velocity will become zero at time, \(0 = 1 - 6 {t_0}\;{\rm{or}}\;{t_0} = \frac{1}{6}s,\) Since the given time \(t = 1\,s\) is greater than \({t_0} = \frac{1}{6}s,\) \({\rm{distance}} > \left| {{\rm{displacement}}} \right|\) Displacement \(S = {x_f} - {x_i}\) \( = (2 + 1 - 3) - (2 + 0 - 0) = - 2\;m\) Distance \(d = \left| {{S_0} - {t_0}} \right| + \left| {{S_t}_{ - {t_0}}} \right|\) \( = \frac{{{u^2}}}{{2\left| a \right|}} + \frac{1}{2}\left| a \right|{\left( {t - {t_0}} \right)^2}\) Comparing \(v = 1 - 6t\) with \(v = u + ut,\) we have \(u = 1\;m{\rm{/}}s\,{\rm{and}}\,a = - 6\;m{\rm{/}}{s^2}\) \(\therefore {\rm{Distance}} = \frac{{{{\left( 1 \right)}^2}}}{{2 \times 6}} + \frac{1}{2} \times 6 \times {\left( {1 - \frac{1}{6}} \right)^2}\) \( = 2.1\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362249
Assertion : The slope of displacement-time of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason : Slope of displacement-time graph = Velocity of the body.
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:
Slope of displacement-time measures velocity of an object. So option (1) is correct
PHXI03:MOTION IN A STRAIGHT LINE
362250
A wheel of radius 1 \(m\) rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
1 \(2\pi \)
2 \(\sqrt 2 \pi \)
3 \(\sqrt {{\pi ^2} + 4} \)
4 \(\pi \)
Explanation:
Horizontal distance by the wheel in half revolution \( = \pi R\) So the displacement of the point which was initially in contact with ground \( = AA' = \sqrt {{{\left( {\pi R} \right)}^2} + {{\left( {2R} \right)}^2}} \) \( = R\sqrt {{\pi ^2} + 4} \) \( = \sqrt {{\pi ^2} + 4} \; ({\rm{as}} \, R = 1 \, m)\)
PHXI03:MOTION IN A STRAIGHT LINE
362251
A person moves towards east for \(4 \, m\), then towards north for \(3 \, m\) and then moves vertically up by \(5 \, m\). What is his distance now from the starting point?
1 \(5\sqrt 2 \, m\)
2 \(5 \, m\)
3 \(10 \, m\)
4 \(20 \, m\)
Explanation:
Distance from starting point \(\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = 5\sqrt 2 \,m\)
362248
The position of a particle along \(x\)-axis at time \(t\) is given by \(x = 2 + t - 3{t^2}\). The displacement and the distance travelled in the interval \(t = 0\,{\rm{and}}\,t = 1\) are respectively
1 \(2,2\)
2 \( - 2,2.5\)
3 \(0,2\)
4 \( - 2,2.1\)
Explanation:
\(x = 2 + t - 3{t^2},\;v = \frac{{dx}}{{dt}} = 1 - 6t,\) velocity will become zero at time, \(0 = 1 - 6 {t_0}\;{\rm{or}}\;{t_0} = \frac{1}{6}s,\) Since the given time \(t = 1\,s\) is greater than \({t_0} = \frac{1}{6}s,\) \({\rm{distance}} > \left| {{\rm{displacement}}} \right|\) Displacement \(S = {x_f} - {x_i}\) \( = (2 + 1 - 3) - (2 + 0 - 0) = - 2\;m\) Distance \(d = \left| {{S_0} - {t_0}} \right| + \left| {{S_t}_{ - {t_0}}} \right|\) \( = \frac{{{u^2}}}{{2\left| a \right|}} + \frac{1}{2}\left| a \right|{\left( {t - {t_0}} \right)^2}\) Comparing \(v = 1 - 6t\) with \(v = u + ut,\) we have \(u = 1\;m{\rm{/}}s\,{\rm{and}}\,a = - 6\;m{\rm{/}}{s^2}\) \(\therefore {\rm{Distance}} = \frac{{{{\left( 1 \right)}^2}}}{{2 \times 6}} + \frac{1}{2} \times 6 \times {\left( {1 - \frac{1}{6}} \right)^2}\) \( = 2.1\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362249
Assertion : The slope of displacement-time of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason : Slope of displacement-time graph = Velocity of the body.
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:
Slope of displacement-time measures velocity of an object. So option (1) is correct
PHXI03:MOTION IN A STRAIGHT LINE
362250
A wheel of radius 1 \(m\) rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
1 \(2\pi \)
2 \(\sqrt 2 \pi \)
3 \(\sqrt {{\pi ^2} + 4} \)
4 \(\pi \)
Explanation:
Horizontal distance by the wheel in half revolution \( = \pi R\) So the displacement of the point which was initially in contact with ground \( = AA' = \sqrt {{{\left( {\pi R} \right)}^2} + {{\left( {2R} \right)}^2}} \) \( = R\sqrt {{\pi ^2} + 4} \) \( = \sqrt {{\pi ^2} + 4} \; ({\rm{as}} \, R = 1 \, m)\)
PHXI03:MOTION IN A STRAIGHT LINE
362251
A person moves towards east for \(4 \, m\), then towards north for \(3 \, m\) and then moves vertically up by \(5 \, m\). What is his distance now from the starting point?
1 \(5\sqrt 2 \, m\)
2 \(5 \, m\)
3 \(10 \, m\)
4 \(20 \, m\)
Explanation:
Distance from starting point \(\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = 5\sqrt 2 \,m\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
PHXI03:MOTION IN A STRAIGHT LINE
362248
The position of a particle along \(x\)-axis at time \(t\) is given by \(x = 2 + t - 3{t^2}\). The displacement and the distance travelled in the interval \(t = 0\,{\rm{and}}\,t = 1\) are respectively
1 \(2,2\)
2 \( - 2,2.5\)
3 \(0,2\)
4 \( - 2,2.1\)
Explanation:
\(x = 2 + t - 3{t^2},\;v = \frac{{dx}}{{dt}} = 1 - 6t,\) velocity will become zero at time, \(0 = 1 - 6 {t_0}\;{\rm{or}}\;{t_0} = \frac{1}{6}s,\) Since the given time \(t = 1\,s\) is greater than \({t_0} = \frac{1}{6}s,\) \({\rm{distance}} > \left| {{\rm{displacement}}} \right|\) Displacement \(S = {x_f} - {x_i}\) \( = (2 + 1 - 3) - (2 + 0 - 0) = - 2\;m\) Distance \(d = \left| {{S_0} - {t_0}} \right| + \left| {{S_t}_{ - {t_0}}} \right|\) \( = \frac{{{u^2}}}{{2\left| a \right|}} + \frac{1}{2}\left| a \right|{\left( {t - {t_0}} \right)^2}\) Comparing \(v = 1 - 6t\) with \(v = u + ut,\) we have \(u = 1\;m{\rm{/}}s\,{\rm{and}}\,a = - 6\;m{\rm{/}}{s^2}\) \(\therefore {\rm{Distance}} = \frac{{{{\left( 1 \right)}^2}}}{{2 \times 6}} + \frac{1}{2} \times 6 \times {\left( {1 - \frac{1}{6}} \right)^2}\) \( = 2.1\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362249
Assertion : The slope of displacement-time of a body moving with high velocity is steeper than the slope of displacement-time graph of a body with low velocity. Reason : Slope of displacement-time graph = Velocity of the body.
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:
Slope of displacement-time measures velocity of an object. So option (1) is correct
PHXI03:MOTION IN A STRAIGHT LINE
362250
A wheel of radius 1 \(m\) rolls forward half a revolution on a horizontal ground. The magnitude of the displacement of the point of the wheel initially in contact with the ground is
1 \(2\pi \)
2 \(\sqrt 2 \pi \)
3 \(\sqrt {{\pi ^2} + 4} \)
4 \(\pi \)
Explanation:
Horizontal distance by the wheel in half revolution \( = \pi R\) So the displacement of the point which was initially in contact with ground \( = AA' = \sqrt {{{\left( {\pi R} \right)}^2} + {{\left( {2R} \right)}^2}} \) \( = R\sqrt {{\pi ^2} + 4} \) \( = \sqrt {{\pi ^2} + 4} \; ({\rm{as}} \, R = 1 \, m)\)
PHXI03:MOTION IN A STRAIGHT LINE
362251
A person moves towards east for \(4 \, m\), then towards north for \(3 \, m\) and then moves vertically up by \(5 \, m\). What is his distance now from the starting point?
1 \(5\sqrt 2 \, m\)
2 \(5 \, m\)
3 \(10 \, m\)
4 \(20 \, m\)
Explanation:
Distance from starting point \(\sqrt {{{\left( 3 \right)}^2} + {{\left( 4 \right)}^2} + {{\left( 5 \right)}^2}} = 5\sqrt 2 \,m\)
