NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI03:MOTION IN A STRAIGHT LINE
362141
A particle moves a distance \(x\) in time \(t\) according to the equation \(x = {(t + 5)^{ - 1}}\). The acceleration of particle is proportional to
1 \({({\rm{distance}})^{ - 2}}\)
2 \({({\rm{distance}})^2}\)
3 \({({\rm{velocity}})^{3/2}}\)
4 \({({\rm{velocity}})^{2/3}}\)
Explanation:
Given, Distance \(x = {(t + 5)^{ - 1}}\) (1) Differentiating Eq. (1) w.r.t. \(t\), we get \(\frac{{dx}}{{dt}} = (v) = \frac{{ - 1}}{{{{(t + 5)}^2}}}\) (2) Again, differentiating Eq. (2) w.r.t. \(t\), we get \(\frac{{{d^2}x}}{{d{t^2}}} = (a) = \frac{2}{{{{(t + 5)}^3}}}\) (3) Comparing Eqs (2) and (3) , we get \(a\;\alpha \,{v^{3/2}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362142
A particle is initially at rest, it is subjected to a linear acceleration \(a\) as shown in figure. The maximum speed attained by the particle is
1 \(605\,m/s\)
2 \(55\,m/s\)
3 \(110\,m/s\)
4 \(550\,m/s\)
Explanation:
Throughout the motion the body is under acceleration but with decreasing \(a\) value. The velocity reaches maximum at the end \({v_{\max }} = \Delta v = {\rm{Area}} = \frac{1}{2} \times 10 \times 11 = 55 \, m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362143
The area of the acceleration-displacement curve of a body gives
1 impulse
2 change in momentum per unit mass
3 change in kinetic energy per unit mass
4 total change in energy
Explanation:
Area of acceleration-displacement curve gives change in kinetic energy per unit mass. \(\frac{1}{2}m\left( {{v^2} - {u^2}} \right) = Fs = \frac{{mdv}}{{dt}} \times s\) \(\therefore \quad \frac{{{\rm{Change}}\,\,\,{\rm{in}}\,\,{\rm{kinetic}}\,\,{\rm{energy}}}}{{{\rm{Mass}}}}\) \( = \frac{{dv}}{{dt}} \times s\)
PHXI03:MOTION IN A STRAIGHT LINE
362144
A car accelerates from rest at a constant for some time, after which it decelerates at a constant rate \(\beta \) and comes to rest. If the total time elapsed is \(t\), then the distance travelled by the car is
362141
A particle moves a distance \(x\) in time \(t\) according to the equation \(x = {(t + 5)^{ - 1}}\). The acceleration of particle is proportional to
1 \({({\rm{distance}})^{ - 2}}\)
2 \({({\rm{distance}})^2}\)
3 \({({\rm{velocity}})^{3/2}}\)
4 \({({\rm{velocity}})^{2/3}}\)
Explanation:
Given, Distance \(x = {(t + 5)^{ - 1}}\) (1) Differentiating Eq. (1) w.r.t. \(t\), we get \(\frac{{dx}}{{dt}} = (v) = \frac{{ - 1}}{{{{(t + 5)}^2}}}\) (2) Again, differentiating Eq. (2) w.r.t. \(t\), we get \(\frac{{{d^2}x}}{{d{t^2}}} = (a) = \frac{2}{{{{(t + 5)}^3}}}\) (3) Comparing Eqs (2) and (3) , we get \(a\;\alpha \,{v^{3/2}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362142
A particle is initially at rest, it is subjected to a linear acceleration \(a\) as shown in figure. The maximum speed attained by the particle is
1 \(605\,m/s\)
2 \(55\,m/s\)
3 \(110\,m/s\)
4 \(550\,m/s\)
Explanation:
Throughout the motion the body is under acceleration but with decreasing \(a\) value. The velocity reaches maximum at the end \({v_{\max }} = \Delta v = {\rm{Area}} = \frac{1}{2} \times 10 \times 11 = 55 \, m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362143
The area of the acceleration-displacement curve of a body gives
1 impulse
2 change in momentum per unit mass
3 change in kinetic energy per unit mass
4 total change in energy
Explanation:
Area of acceleration-displacement curve gives change in kinetic energy per unit mass. \(\frac{1}{2}m\left( {{v^2} - {u^2}} \right) = Fs = \frac{{mdv}}{{dt}} \times s\) \(\therefore \quad \frac{{{\rm{Change}}\,\,\,{\rm{in}}\,\,{\rm{kinetic}}\,\,{\rm{energy}}}}{{{\rm{Mass}}}}\) \( = \frac{{dv}}{{dt}} \times s\)
PHXI03:MOTION IN A STRAIGHT LINE
362144
A car accelerates from rest at a constant for some time, after which it decelerates at a constant rate \(\beta \) and comes to rest. If the total time elapsed is \(t\), then the distance travelled by the car is
362141
A particle moves a distance \(x\) in time \(t\) according to the equation \(x = {(t + 5)^{ - 1}}\). The acceleration of particle is proportional to
1 \({({\rm{distance}})^{ - 2}}\)
2 \({({\rm{distance}})^2}\)
3 \({({\rm{velocity}})^{3/2}}\)
4 \({({\rm{velocity}})^{2/3}}\)
Explanation:
Given, Distance \(x = {(t + 5)^{ - 1}}\) (1) Differentiating Eq. (1) w.r.t. \(t\), we get \(\frac{{dx}}{{dt}} = (v) = \frac{{ - 1}}{{{{(t + 5)}^2}}}\) (2) Again, differentiating Eq. (2) w.r.t. \(t\), we get \(\frac{{{d^2}x}}{{d{t^2}}} = (a) = \frac{2}{{{{(t + 5)}^3}}}\) (3) Comparing Eqs (2) and (3) , we get \(a\;\alpha \,{v^{3/2}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362142
A particle is initially at rest, it is subjected to a linear acceleration \(a\) as shown in figure. The maximum speed attained by the particle is
1 \(605\,m/s\)
2 \(55\,m/s\)
3 \(110\,m/s\)
4 \(550\,m/s\)
Explanation:
Throughout the motion the body is under acceleration but with decreasing \(a\) value. The velocity reaches maximum at the end \({v_{\max }} = \Delta v = {\rm{Area}} = \frac{1}{2} \times 10 \times 11 = 55 \, m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362143
The area of the acceleration-displacement curve of a body gives
1 impulse
2 change in momentum per unit mass
3 change in kinetic energy per unit mass
4 total change in energy
Explanation:
Area of acceleration-displacement curve gives change in kinetic energy per unit mass. \(\frac{1}{2}m\left( {{v^2} - {u^2}} \right) = Fs = \frac{{mdv}}{{dt}} \times s\) \(\therefore \quad \frac{{{\rm{Change}}\,\,\,{\rm{in}}\,\,{\rm{kinetic}}\,\,{\rm{energy}}}}{{{\rm{Mass}}}}\) \( = \frac{{dv}}{{dt}} \times s\)
PHXI03:MOTION IN A STRAIGHT LINE
362144
A car accelerates from rest at a constant for some time, after which it decelerates at a constant rate \(\beta \) and comes to rest. If the total time elapsed is \(t\), then the distance travelled by the car is
362141
A particle moves a distance \(x\) in time \(t\) according to the equation \(x = {(t + 5)^{ - 1}}\). The acceleration of particle is proportional to
1 \({({\rm{distance}})^{ - 2}}\)
2 \({({\rm{distance}})^2}\)
3 \({({\rm{velocity}})^{3/2}}\)
4 \({({\rm{velocity}})^{2/3}}\)
Explanation:
Given, Distance \(x = {(t + 5)^{ - 1}}\) (1) Differentiating Eq. (1) w.r.t. \(t\), we get \(\frac{{dx}}{{dt}} = (v) = \frac{{ - 1}}{{{{(t + 5)}^2}}}\) (2) Again, differentiating Eq. (2) w.r.t. \(t\), we get \(\frac{{{d^2}x}}{{d{t^2}}} = (a) = \frac{2}{{{{(t + 5)}^3}}}\) (3) Comparing Eqs (2) and (3) , we get \(a\;\alpha \,{v^{3/2}}\)
PHXI03:MOTION IN A STRAIGHT LINE
362142
A particle is initially at rest, it is subjected to a linear acceleration \(a\) as shown in figure. The maximum speed attained by the particle is
1 \(605\,m/s\)
2 \(55\,m/s\)
3 \(110\,m/s\)
4 \(550\,m/s\)
Explanation:
Throughout the motion the body is under acceleration but with decreasing \(a\) value. The velocity reaches maximum at the end \({v_{\max }} = \Delta v = {\rm{Area}} = \frac{1}{2} \times 10 \times 11 = 55 \, m/s\)
PHXI03:MOTION IN A STRAIGHT LINE
362143
The area of the acceleration-displacement curve of a body gives
1 impulse
2 change in momentum per unit mass
3 change in kinetic energy per unit mass
4 total change in energy
Explanation:
Area of acceleration-displacement curve gives change in kinetic energy per unit mass. \(\frac{1}{2}m\left( {{v^2} - {u^2}} \right) = Fs = \frac{{mdv}}{{dt}} \times s\) \(\therefore \quad \frac{{{\rm{Change}}\,\,\,{\rm{in}}\,\,{\rm{kinetic}}\,\,{\rm{energy}}}}{{{\rm{Mass}}}}\) \( = \frac{{dv}}{{dt}} \times s\)
PHXI03:MOTION IN A STRAIGHT LINE
362144
A car accelerates from rest at a constant for some time, after which it decelerates at a constant rate \(\beta \) and comes to rest. If the total time elapsed is \(t\), then the distance travelled by the car is
