362162
Assertion : A body having non-zero acceleration can have a constant velocity. Reason : Acceleration is the 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:
Acceleration is the rate of change of velocity, i.e., \(a = \frac{{dv}}{{dt}}\) So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362163
The relation between time and distance is \(t = \alpha {x^2} + \beta x,\) where \(\alpha \) and \(\beta \) are constants. The retardation is
1 \(2\alpha \beta {v^3}\)
2 \(2\alpha {v^3}\)
3 \(2{\beta ^2}{v^3}\)
4 \(2\beta {v^3}\)
Explanation:
\(\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}\) \(\because a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}\) \(a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{\left( {2\alpha x + \beta } \right)}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}\) \(\therefore \quad {\rm{Retardation}} = 2\alpha {v^3}.\)
PHXI03:MOTION IN A STRAIGHT LINE
362164
Study the following graphs: The particle has negative acceleration
1 In graph (ii)
2 In graph (i)
3 In graph (iii)
4 In graph (iv)
Explanation:
For the graphs (i) and (iv), slope is constant, hence the velocity is constant and hence acceleration is zero. For the graph (iii), the particle’s velocity first decreases and then increases. It means negative acceleration is involved in this motion. For graph (ii) velocity increases and hence it is an accelerated motion.
PHXI03:MOTION IN A STRAIGHT LINE
362165
The displacement of a particle is given by \(y = a + bt + c{t^2} - d{t^4}\). The initial velocity and acceleration are respectively
1 \(b, - 4d\)
2 \( - b, - 2c\)
3 \(b,2c\)
4 \(2c, - 4d\)
Explanation:
\(y = a + bt + c{t^2} - d{t^4}\) \(\therefore v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}\) and \(a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}\) Hence, at \(t = 0,{v_{initial}} = b\;{\rm{and}}\;{a_{initial}} = 2c\)
PHXI03:MOTION IN A STRAIGHT LINE
362166
The velocity-time graph of moving object is given in the figure. Distance travelled by the body in the interval of time in which the acceleration is maximum is
1 \(320\,m\)
2 \(150\,m\)
3 \(250\,m\)
4 \(255\,m\)
Explanation:
Maximum acceleration is from \(B\) to \(C\). \(\therefore \quad {a_{\max }}{\rm{ }} = \) slope of \(BC\) curve \( = \frac{{\Delta v}}{{\Delta t}}\) \( = \frac{{40 - 10}}{{40 - 30}} = 3\;m\;{s^{ - 2}}\;\) \({s_{\max }}{\rm{ }} = \frac{{{v^2} - {u^2}}}{{2{a_{\max }}}} = \frac{{{{40}^2} - {{10}^2}}}{{2 \times 3}} = 250\;m\)
362162
Assertion : A body having non-zero acceleration can have a constant velocity. Reason : Acceleration is the 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:
Acceleration is the rate of change of velocity, i.e., \(a = \frac{{dv}}{{dt}}\) So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362163
The relation between time and distance is \(t = \alpha {x^2} + \beta x,\) where \(\alpha \) and \(\beta \) are constants. The retardation is
1 \(2\alpha \beta {v^3}\)
2 \(2\alpha {v^3}\)
3 \(2{\beta ^2}{v^3}\)
4 \(2\beta {v^3}\)
Explanation:
\(\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}\) \(\because a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}\) \(a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{\left( {2\alpha x + \beta } \right)}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}\) \(\therefore \quad {\rm{Retardation}} = 2\alpha {v^3}.\)
PHXI03:MOTION IN A STRAIGHT LINE
362164
Study the following graphs: The particle has negative acceleration
1 In graph (ii)
2 In graph (i)
3 In graph (iii)
4 In graph (iv)
Explanation:
For the graphs (i) and (iv), slope is constant, hence the velocity is constant and hence acceleration is zero. For the graph (iii), the particle’s velocity first decreases and then increases. It means negative acceleration is involved in this motion. For graph (ii) velocity increases and hence it is an accelerated motion.
PHXI03:MOTION IN A STRAIGHT LINE
362165
The displacement of a particle is given by \(y = a + bt + c{t^2} - d{t^4}\). The initial velocity and acceleration are respectively
1 \(b, - 4d\)
2 \( - b, - 2c\)
3 \(b,2c\)
4 \(2c, - 4d\)
Explanation:
\(y = a + bt + c{t^2} - d{t^4}\) \(\therefore v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}\) and \(a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}\) Hence, at \(t = 0,{v_{initial}} = b\;{\rm{and}}\;{a_{initial}} = 2c\)
PHXI03:MOTION IN A STRAIGHT LINE
362166
The velocity-time graph of moving object is given in the figure. Distance travelled by the body in the interval of time in which the acceleration is maximum is
1 \(320\,m\)
2 \(150\,m\)
3 \(250\,m\)
4 \(255\,m\)
Explanation:
Maximum acceleration is from \(B\) to \(C\). \(\therefore \quad {a_{\max }}{\rm{ }} = \) slope of \(BC\) curve \( = \frac{{\Delta v}}{{\Delta t}}\) \( = \frac{{40 - 10}}{{40 - 30}} = 3\;m\;{s^{ - 2}}\;\) \({s_{\max }}{\rm{ }} = \frac{{{v^2} - {u^2}}}{{2{a_{\max }}}} = \frac{{{{40}^2} - {{10}^2}}}{{2 \times 3}} = 250\;m\)
362162
Assertion : A body having non-zero acceleration can have a constant velocity. Reason : Acceleration is the 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:
Acceleration is the rate of change of velocity, i.e., \(a = \frac{{dv}}{{dt}}\) So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362163
The relation between time and distance is \(t = \alpha {x^2} + \beta x,\) where \(\alpha \) and \(\beta \) are constants. The retardation is
1 \(2\alpha \beta {v^3}\)
2 \(2\alpha {v^3}\)
3 \(2{\beta ^2}{v^3}\)
4 \(2\beta {v^3}\)
Explanation:
\(\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}\) \(\because a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}\) \(a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{\left( {2\alpha x + \beta } \right)}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}\) \(\therefore \quad {\rm{Retardation}} = 2\alpha {v^3}.\)
PHXI03:MOTION IN A STRAIGHT LINE
362164
Study the following graphs: The particle has negative acceleration
1 In graph (ii)
2 In graph (i)
3 In graph (iii)
4 In graph (iv)
Explanation:
For the graphs (i) and (iv), slope is constant, hence the velocity is constant and hence acceleration is zero. For the graph (iii), the particle’s velocity first decreases and then increases. It means negative acceleration is involved in this motion. For graph (ii) velocity increases and hence it is an accelerated motion.
PHXI03:MOTION IN A STRAIGHT LINE
362165
The displacement of a particle is given by \(y = a + bt + c{t^2} - d{t^4}\). The initial velocity and acceleration are respectively
1 \(b, - 4d\)
2 \( - b, - 2c\)
3 \(b,2c\)
4 \(2c, - 4d\)
Explanation:
\(y = a + bt + c{t^2} - d{t^4}\) \(\therefore v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}\) and \(a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}\) Hence, at \(t = 0,{v_{initial}} = b\;{\rm{and}}\;{a_{initial}} = 2c\)
PHXI03:MOTION IN A STRAIGHT LINE
362166
The velocity-time graph of moving object is given in the figure. Distance travelled by the body in the interval of time in which the acceleration is maximum is
1 \(320\,m\)
2 \(150\,m\)
3 \(250\,m\)
4 \(255\,m\)
Explanation:
Maximum acceleration is from \(B\) to \(C\). \(\therefore \quad {a_{\max }}{\rm{ }} = \) slope of \(BC\) curve \( = \frac{{\Delta v}}{{\Delta t}}\) \( = \frac{{40 - 10}}{{40 - 30}} = 3\;m\;{s^{ - 2}}\;\) \({s_{\max }}{\rm{ }} = \frac{{{v^2} - {u^2}}}{{2{a_{\max }}}} = \frac{{{{40}^2} - {{10}^2}}}{{2 \times 3}} = 250\;m\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI03:MOTION IN A STRAIGHT LINE
362162
Assertion : A body having non-zero acceleration can have a constant velocity. Reason : Acceleration is the 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:
Acceleration is the rate of change of velocity, i.e., \(a = \frac{{dv}}{{dt}}\) So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362163
The relation between time and distance is \(t = \alpha {x^2} + \beta x,\) where \(\alpha \) and \(\beta \) are constants. The retardation is
1 \(2\alpha \beta {v^3}\)
2 \(2\alpha {v^3}\)
3 \(2{\beta ^2}{v^3}\)
4 \(2\beta {v^3}\)
Explanation:
\(\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}\) \(\because a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}\) \(a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{\left( {2\alpha x + \beta } \right)}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}\) \(\therefore \quad {\rm{Retardation}} = 2\alpha {v^3}.\)
PHXI03:MOTION IN A STRAIGHT LINE
362164
Study the following graphs: The particle has negative acceleration
1 In graph (ii)
2 In graph (i)
3 In graph (iii)
4 In graph (iv)
Explanation:
For the graphs (i) and (iv), slope is constant, hence the velocity is constant and hence acceleration is zero. For the graph (iii), the particle’s velocity first decreases and then increases. It means negative acceleration is involved in this motion. For graph (ii) velocity increases and hence it is an accelerated motion.
PHXI03:MOTION IN A STRAIGHT LINE
362165
The displacement of a particle is given by \(y = a + bt + c{t^2} - d{t^4}\). The initial velocity and acceleration are respectively
1 \(b, - 4d\)
2 \( - b, - 2c\)
3 \(b,2c\)
4 \(2c, - 4d\)
Explanation:
\(y = a + bt + c{t^2} - d{t^4}\) \(\therefore v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}\) and \(a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}\) Hence, at \(t = 0,{v_{initial}} = b\;{\rm{and}}\;{a_{initial}} = 2c\)
PHXI03:MOTION IN A STRAIGHT LINE
362166
The velocity-time graph of moving object is given in the figure. Distance travelled by the body in the interval of time in which the acceleration is maximum is
1 \(320\,m\)
2 \(150\,m\)
3 \(250\,m\)
4 \(255\,m\)
Explanation:
Maximum acceleration is from \(B\) to \(C\). \(\therefore \quad {a_{\max }}{\rm{ }} = \) slope of \(BC\) curve \( = \frac{{\Delta v}}{{\Delta t}}\) \( = \frac{{40 - 10}}{{40 - 30}} = 3\;m\;{s^{ - 2}}\;\) \({s_{\max }}{\rm{ }} = \frac{{{v^2} - {u^2}}}{{2{a_{\max }}}} = \frac{{{{40}^2} - {{10}^2}}}{{2 \times 3}} = 250\;m\)
362162
Assertion : A body having non-zero acceleration can have a constant velocity. Reason : Acceleration is the 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:
Acceleration is the rate of change of velocity, i.e., \(a = \frac{{dv}}{{dt}}\) So option (4) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362163
The relation between time and distance is \(t = \alpha {x^2} + \beta x,\) where \(\alpha \) and \(\beta \) are constants. The retardation is
1 \(2\alpha \beta {v^3}\)
2 \(2\alpha {v^3}\)
3 \(2{\beta ^2}{v^3}\)
4 \(2\beta {v^3}\)
Explanation:
\(\frac{{dt}}{{dx}} = 2\alpha x + \beta \Rightarrow v = \frac{1}{{2\alpha x + \beta }}\) \(\because a = \frac{{dv}}{{dt}} = \frac{{dv}}{{dx}}.\frac{{dx}}{{dt}}\) \(a = v\frac{{dv}}{{dx}} = \frac{{ - v.2\alpha }}{{{{\left( {2\alpha x + \beta } \right)}^2}}} = - 2\alpha .v.{v^2} = - 2\alpha {v^3}\) \(\therefore \quad {\rm{Retardation}} = 2\alpha {v^3}.\)
PHXI03:MOTION IN A STRAIGHT LINE
362164
Study the following graphs: The particle has negative acceleration
1 In graph (ii)
2 In graph (i)
3 In graph (iii)
4 In graph (iv)
Explanation:
For the graphs (i) and (iv), slope is constant, hence the velocity is constant and hence acceleration is zero. For the graph (iii), the particle’s velocity first decreases and then increases. It means negative acceleration is involved in this motion. For graph (ii) velocity increases and hence it is an accelerated motion.
PHXI03:MOTION IN A STRAIGHT LINE
362165
The displacement of a particle is given by \(y = a + bt + c{t^2} - d{t^4}\). The initial velocity and acceleration are respectively
1 \(b, - 4d\)
2 \( - b, - 2c\)
3 \(b,2c\)
4 \(2c, - 4d\)
Explanation:
\(y = a + bt + c{t^2} - d{t^4}\) \(\therefore v = \frac{{dy}}{{dt}} = b + 2ct - 4d{t^3}\) and \(a = \frac{{dv}}{{dt}} = 2c - 12d{t^2}\) Hence, at \(t = 0,{v_{initial}} = b\;{\rm{and}}\;{a_{initial}} = 2c\)
PHXI03:MOTION IN A STRAIGHT LINE
362166
The velocity-time graph of moving object is given in the figure. Distance travelled by the body in the interval of time in which the acceleration is maximum is
1 \(320\,m\)
2 \(150\,m\)
3 \(250\,m\)
4 \(255\,m\)
Explanation:
Maximum acceleration is from \(B\) to \(C\). \(\therefore \quad {a_{\max }}{\rm{ }} = \) slope of \(BC\) curve \( = \frac{{\Delta v}}{{\Delta t}}\) \( = \frac{{40 - 10}}{{40 - 30}} = 3\;m\;{s^{ - 2}}\;\) \({s_{\max }}{\rm{ }} = \frac{{{v^2} - {u^2}}}{{2{a_{\max }}}} = \frac{{{{40}^2} - {{10}^2}}}{{2 \times 3}} = 250\;m\)
