362276
The displacement of the particle starting from rest (at \(t = 0\)) is given by \(s = 6{t^2} - {t^3}\). The time in second at which the particle will attain zero velocity again, is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(8\)
Explanation:
\(v = \frac{{ds}}{{dt}} = 12t - 3{t^2}\) Velocity is zero for \(t = 0\) and \(t = 4s\)
PHXI03:MOTION IN A STRAIGHT LINE
362277
The displacement-time graph of a moving particle is shown in figure. The instantaneous velocity of the particle is negative at the point.
1 \(F\)
2 \(D\)
3 \(E\)
4 \(C\)
Explanation:
The slope of the graph is negative at this point.
PHXI03:MOTION IN A STRAIGHT LINE
362278
Assertion : The instantaneous velocity does not depend on instantaneous position vector. Reason : The instantaneous velocity and average velocity of a particle are always same.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Instantaneous velocity depends solely on the rate of change of position with respect to time, and not on the position itself. Instantaneous velocity and average velocity are not always the same. Average velocity is calculated over a finite time interval, while instantaneous velocity refers to the velocity at a single point in time. (i.e, at an instant) So correct option is (3).
PHXI03:MOTION IN A STRAIGHT LINE
362279
The engine of a train passes an electric pole with a velocity ‘\(u\)’ and the last compartment of the train crosses the same pole with a velocity \(v\). Then the velocity with which the mid-point of the train passes the pole is
1 \(u\)
2 \(v\)
3 \(\frac{{u + v}}{2}\)
4 \(\sqrt {\frac{{{u^2} + {v^2}}}{2}} \)
Explanation:
Let \(l\) be the total length of the train and \(v'\) be the velocity of the midpoint of train \(v{'^2} - {u^2} = 2a\frac{l}{2}\) (1) \(v' - u{'^2} = 2a.l\) (2) From (1) and (2) we get \(v' = \sqrt {\frac{{{v^2} + {u^2}}}{2}} \)
362276
The displacement of the particle starting from rest (at \(t = 0\)) is given by \(s = 6{t^2} - {t^3}\). The time in second at which the particle will attain zero velocity again, is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(8\)
Explanation:
\(v = \frac{{ds}}{{dt}} = 12t - 3{t^2}\) Velocity is zero for \(t = 0\) and \(t = 4s\)
PHXI03:MOTION IN A STRAIGHT LINE
362277
The displacement-time graph of a moving particle is shown in figure. The instantaneous velocity of the particle is negative at the point.
1 \(F\)
2 \(D\)
3 \(E\)
4 \(C\)
Explanation:
The slope of the graph is negative at this point.
PHXI03:MOTION IN A STRAIGHT LINE
362278
Assertion : The instantaneous velocity does not depend on instantaneous position vector. Reason : The instantaneous velocity and average velocity of a particle are always same.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Instantaneous velocity depends solely on the rate of change of position with respect to time, and not on the position itself. Instantaneous velocity and average velocity are not always the same. Average velocity is calculated over a finite time interval, while instantaneous velocity refers to the velocity at a single point in time. (i.e, at an instant) So correct option is (3).
PHXI03:MOTION IN A STRAIGHT LINE
362279
The engine of a train passes an electric pole with a velocity ‘\(u\)’ and the last compartment of the train crosses the same pole with a velocity \(v\). Then the velocity with which the mid-point of the train passes the pole is
1 \(u\)
2 \(v\)
3 \(\frac{{u + v}}{2}\)
4 \(\sqrt {\frac{{{u^2} + {v^2}}}{2}} \)
Explanation:
Let \(l\) be the total length of the train and \(v'\) be the velocity of the midpoint of train \(v{'^2} - {u^2} = 2a\frac{l}{2}\) (1) \(v' - u{'^2} = 2a.l\) (2) From (1) and (2) we get \(v' = \sqrt {\frac{{{v^2} + {u^2}}}{2}} \)
362276
The displacement of the particle starting from rest (at \(t = 0\)) is given by \(s = 6{t^2} - {t^3}\). The time in second at which the particle will attain zero velocity again, is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(8\)
Explanation:
\(v = \frac{{ds}}{{dt}} = 12t - 3{t^2}\) Velocity is zero for \(t = 0\) and \(t = 4s\)
PHXI03:MOTION IN A STRAIGHT LINE
362277
The displacement-time graph of a moving particle is shown in figure. The instantaneous velocity of the particle is negative at the point.
1 \(F\)
2 \(D\)
3 \(E\)
4 \(C\)
Explanation:
The slope of the graph is negative at this point.
PHXI03:MOTION IN A STRAIGHT LINE
362278
Assertion : The instantaneous velocity does not depend on instantaneous position vector. Reason : The instantaneous velocity and average velocity of a particle are always same.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Instantaneous velocity depends solely on the rate of change of position with respect to time, and not on the position itself. Instantaneous velocity and average velocity are not always the same. Average velocity is calculated over a finite time interval, while instantaneous velocity refers to the velocity at a single point in time. (i.e, at an instant) So correct option is (3).
PHXI03:MOTION IN A STRAIGHT LINE
362279
The engine of a train passes an electric pole with a velocity ‘\(u\)’ and the last compartment of the train crosses the same pole with a velocity \(v\). Then the velocity with which the mid-point of the train passes the pole is
1 \(u\)
2 \(v\)
3 \(\frac{{u + v}}{2}\)
4 \(\sqrt {\frac{{{u^2} + {v^2}}}{2}} \)
Explanation:
Let \(l\) be the total length of the train and \(v'\) be the velocity of the midpoint of train \(v{'^2} - {u^2} = 2a\frac{l}{2}\) (1) \(v' - u{'^2} = 2a.l\) (2) From (1) and (2) we get \(v' = \sqrt {\frac{{{v^2} + {u^2}}}{2}} \)
362276
The displacement of the particle starting from rest (at \(t = 0\)) is given by \(s = 6{t^2} - {t^3}\). The time in second at which the particle will attain zero velocity again, is
1 \(2\)
2 \(4\)
3 \(6\)
4 \(8\)
Explanation:
\(v = \frac{{ds}}{{dt}} = 12t - 3{t^2}\) Velocity is zero for \(t = 0\) and \(t = 4s\)
PHXI03:MOTION IN A STRAIGHT LINE
362277
The displacement-time graph of a moving particle is shown in figure. The instantaneous velocity of the particle is negative at the point.
1 \(F\)
2 \(D\)
3 \(E\)
4 \(C\)
Explanation:
The slope of the graph is negative at this point.
PHXI03:MOTION IN A STRAIGHT LINE
362278
Assertion : The instantaneous velocity does not depend on instantaneous position vector. Reason : The instantaneous velocity and average velocity of a particle are always same.
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but reason is correct.
Explanation:
Instantaneous velocity depends solely on the rate of change of position with respect to time, and not on the position itself. Instantaneous velocity and average velocity are not always the same. Average velocity is calculated over a finite time interval, while instantaneous velocity refers to the velocity at a single point in time. (i.e, at an instant) So correct option is (3).
PHXI03:MOTION IN A STRAIGHT LINE
362279
The engine of a train passes an electric pole with a velocity ‘\(u\)’ and the last compartment of the train crosses the same pole with a velocity \(v\). Then the velocity with which the mid-point of the train passes the pole is
1 \(u\)
2 \(v\)
3 \(\frac{{u + v}}{2}\)
4 \(\sqrt {\frac{{{u^2} + {v^2}}}{2}} \)
Explanation:
Let \(l\) be the total length of the train and \(v'\) be the velocity of the midpoint of train \(v{'^2} - {u^2} = 2a\frac{l}{2}\) (1) \(v' - u{'^2} = 2a.l\) (2) From (1) and (2) we get \(v' = \sqrt {\frac{{{v^2} + {u^2}}}{2}} \)
