141255
The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
1 The total distance travelled by the particle is zero
2 The total displacement of the particle is zero
3 The maximum acceleration of the particle is \(\frac{1}{2} \mathrm{~ms}^{-2}\)
4 The total distance travelled by the particle at the end of \(10 \mathrm{~s}\) is \(100 \mathrm{~m}\)
5 At the fifth second, the acceleration of the particle is \(2 \mathrm{~ms}^{-2}\)
Explanation:
B \(\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\int \mathrm{ds}=\int \mathrm{vdt}\) \(\Delta \mathrm{s}=\) Area under v-t curve \(\Delta \mathrm{s}=\mathrm{d}_{1}+\mathrm{d}_{2}\) \(\Delta \mathrm{s}=\frac{1}{2} \times 10 \times 10-\frac{1}{2} \times 10 \times 10\) \(\therefore \quad \Delta \mathrm{s}=0\)
Kerala CEE -2018
Motion in One Dimensions
141257
A car starts from rest with a constant acceleration of \(5 \mathrm{~ms}^{-2}\). At the same time a bus travelling with uniform velocity of \(50 \mathrm{~ms}^{-1}\) overtakes and passes the car. The distance through which the car again overtakes the bus is:
1 \(250 \mathrm{~m}\)
2 \(500 \mathrm{~m}\)
3 \(750 \mathrm{~m}\)
4 \(1000 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{s}=50 \mathrm{t}, \mathrm{u}=0, \mathrm{a}=5 \mathrm{~m} / \mathrm{s}^{2}\) Distance covered by bus will be equal to distance covered by car. \(\because \quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(50 \mathrm{t}=0 \times \mathrm{t}+\frac{1}{2} 5 \mathrm{t}^{2}\) \(\mathrm{t}=20 \mathrm{sec}\) Distance \(=\) speed \(\times\) time \(=50 \times 20=1000 \mathrm{~m}\)
Assam CEE-2017
Motion in One Dimensions
141258
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
141259
The displacement of a particle moving along \(x\) axis versus time is given in the figure below: The average velocity \(\left(\mathrm{V}_{\text {avg }}\right)\) of the particle in the first four seconds and the velocity \(V\) of it at \(t=4 s\) are
D Let us consider As \(\mathrm{t}=0\), particle is at a position \(5 \mathrm{~m}\), and at \(\mathrm{t}=4 \mathrm{~s}\) particle is at a position \(1 \mathrm{~m}\). Displacement of particle \((\mathrm{s})=1-5=-4 \mathrm{~m}\) time taken \((\mathrm{t})=4 \mathrm{~s}\) Average velocity \(\left(\mathrm{V}_{\text {avg }}\right)=\frac{\mathrm{s}}{\mathrm{t}}=\frac{-4}{4}=-1 \mathrm{~m} / \mathrm{s}\) At a slope of displacement time graph gives the velocity of particle at that instant. slope of the graph is zero at \(t=4 \mathrm{~s}\) then, the velocity is zero at \(t=4 \mathrm{~s}\) Hence, \(V_{a v}=-1 \mathrm{~m} / \mathrm{s}, V=0 \mathrm{~m} / \mathrm{s}\)
141255
The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
1 The total distance travelled by the particle is zero
2 The total displacement of the particle is zero
3 The maximum acceleration of the particle is \(\frac{1}{2} \mathrm{~ms}^{-2}\)
4 The total distance travelled by the particle at the end of \(10 \mathrm{~s}\) is \(100 \mathrm{~m}\)
5 At the fifth second, the acceleration of the particle is \(2 \mathrm{~ms}^{-2}\)
Explanation:
B \(\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\int \mathrm{ds}=\int \mathrm{vdt}\) \(\Delta \mathrm{s}=\) Area under v-t curve \(\Delta \mathrm{s}=\mathrm{d}_{1}+\mathrm{d}_{2}\) \(\Delta \mathrm{s}=\frac{1}{2} \times 10 \times 10-\frac{1}{2} \times 10 \times 10\) \(\therefore \quad \Delta \mathrm{s}=0\)
Kerala CEE -2018
Motion in One Dimensions
141257
A car starts from rest with a constant acceleration of \(5 \mathrm{~ms}^{-2}\). At the same time a bus travelling with uniform velocity of \(50 \mathrm{~ms}^{-1}\) overtakes and passes the car. The distance through which the car again overtakes the bus is:
1 \(250 \mathrm{~m}\)
2 \(500 \mathrm{~m}\)
3 \(750 \mathrm{~m}\)
4 \(1000 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{s}=50 \mathrm{t}, \mathrm{u}=0, \mathrm{a}=5 \mathrm{~m} / \mathrm{s}^{2}\) Distance covered by bus will be equal to distance covered by car. \(\because \quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(50 \mathrm{t}=0 \times \mathrm{t}+\frac{1}{2} 5 \mathrm{t}^{2}\) \(\mathrm{t}=20 \mathrm{sec}\) Distance \(=\) speed \(\times\) time \(=50 \times 20=1000 \mathrm{~m}\)
Assam CEE-2017
Motion in One Dimensions
141258
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
141259
The displacement of a particle moving along \(x\) axis versus time is given in the figure below: The average velocity \(\left(\mathrm{V}_{\text {avg }}\right)\) of the particle in the first four seconds and the velocity \(V\) of it at \(t=4 s\) are
D Let us consider As \(\mathrm{t}=0\), particle is at a position \(5 \mathrm{~m}\), and at \(\mathrm{t}=4 \mathrm{~s}\) particle is at a position \(1 \mathrm{~m}\). Displacement of particle \((\mathrm{s})=1-5=-4 \mathrm{~m}\) time taken \((\mathrm{t})=4 \mathrm{~s}\) Average velocity \(\left(\mathrm{V}_{\text {avg }}\right)=\frac{\mathrm{s}}{\mathrm{t}}=\frac{-4}{4}=-1 \mathrm{~m} / \mathrm{s}\) At a slope of displacement time graph gives the velocity of particle at that instant. slope of the graph is zero at \(t=4 \mathrm{~s}\) then, the velocity is zero at \(t=4 \mathrm{~s}\) Hence, \(V_{a v}=-1 \mathrm{~m} / \mathrm{s}, V=0 \mathrm{~m} / \mathrm{s}\)
141255
The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
1 The total distance travelled by the particle is zero
2 The total displacement of the particle is zero
3 The maximum acceleration of the particle is \(\frac{1}{2} \mathrm{~ms}^{-2}\)
4 The total distance travelled by the particle at the end of \(10 \mathrm{~s}\) is \(100 \mathrm{~m}\)
5 At the fifth second, the acceleration of the particle is \(2 \mathrm{~ms}^{-2}\)
Explanation:
B \(\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\int \mathrm{ds}=\int \mathrm{vdt}\) \(\Delta \mathrm{s}=\) Area under v-t curve \(\Delta \mathrm{s}=\mathrm{d}_{1}+\mathrm{d}_{2}\) \(\Delta \mathrm{s}=\frac{1}{2} \times 10 \times 10-\frac{1}{2} \times 10 \times 10\) \(\therefore \quad \Delta \mathrm{s}=0\)
Kerala CEE -2018
Motion in One Dimensions
141257
A car starts from rest with a constant acceleration of \(5 \mathrm{~ms}^{-2}\). At the same time a bus travelling with uniform velocity of \(50 \mathrm{~ms}^{-1}\) overtakes and passes the car. The distance through which the car again overtakes the bus is:
1 \(250 \mathrm{~m}\)
2 \(500 \mathrm{~m}\)
3 \(750 \mathrm{~m}\)
4 \(1000 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{s}=50 \mathrm{t}, \mathrm{u}=0, \mathrm{a}=5 \mathrm{~m} / \mathrm{s}^{2}\) Distance covered by bus will be equal to distance covered by car. \(\because \quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(50 \mathrm{t}=0 \times \mathrm{t}+\frac{1}{2} 5 \mathrm{t}^{2}\) \(\mathrm{t}=20 \mathrm{sec}\) Distance \(=\) speed \(\times\) time \(=50 \times 20=1000 \mathrm{~m}\)
Assam CEE-2017
Motion in One Dimensions
141258
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
141259
The displacement of a particle moving along \(x\) axis versus time is given in the figure below: The average velocity \(\left(\mathrm{V}_{\text {avg }}\right)\) of the particle in the first four seconds and the velocity \(V\) of it at \(t=4 s\) are
D Let us consider As \(\mathrm{t}=0\), particle is at a position \(5 \mathrm{~m}\), and at \(\mathrm{t}=4 \mathrm{~s}\) particle is at a position \(1 \mathrm{~m}\). Displacement of particle \((\mathrm{s})=1-5=-4 \mathrm{~m}\) time taken \((\mathrm{t})=4 \mathrm{~s}\) Average velocity \(\left(\mathrm{V}_{\text {avg }}\right)=\frac{\mathrm{s}}{\mathrm{t}}=\frac{-4}{4}=-1 \mathrm{~m} / \mathrm{s}\) At a slope of displacement time graph gives the velocity of particle at that instant. slope of the graph is zero at \(t=4 \mathrm{~s}\) then, the velocity is zero at \(t=4 \mathrm{~s}\) Hence, \(V_{a v}=-1 \mathrm{~m} / \mathrm{s}, V=0 \mathrm{~m} / \mathrm{s}\)
141255
The one-dimensional motion of a point particle is shown in the figure. Select the correct statement.
1 The total distance travelled by the particle is zero
2 The total displacement of the particle is zero
3 The maximum acceleration of the particle is \(\frac{1}{2} \mathrm{~ms}^{-2}\)
4 The total distance travelled by the particle at the end of \(10 \mathrm{~s}\) is \(100 \mathrm{~m}\)
5 At the fifth second, the acceleration of the particle is \(2 \mathrm{~ms}^{-2}\)
Explanation:
B \(\mathrm{v}=\frac{\mathrm{ds}}{\mathrm{dt}}\) \(\int \mathrm{ds}=\int \mathrm{vdt}\) \(\Delta \mathrm{s}=\) Area under v-t curve \(\Delta \mathrm{s}=\mathrm{d}_{1}+\mathrm{d}_{2}\) \(\Delta \mathrm{s}=\frac{1}{2} \times 10 \times 10-\frac{1}{2} \times 10 \times 10\) \(\therefore \quad \Delta \mathrm{s}=0\)
Kerala CEE -2018
Motion in One Dimensions
141257
A car starts from rest with a constant acceleration of \(5 \mathrm{~ms}^{-2}\). At the same time a bus travelling with uniform velocity of \(50 \mathrm{~ms}^{-1}\) overtakes and passes the car. The distance through which the car again overtakes the bus is:
1 \(250 \mathrm{~m}\)
2 \(500 \mathrm{~m}\)
3 \(750 \mathrm{~m}\)
4 \(1000 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{s}=50 \mathrm{t}, \mathrm{u}=0, \mathrm{a}=5 \mathrm{~m} / \mathrm{s}^{2}\) Distance covered by bus will be equal to distance covered by car. \(\because \quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(50 \mathrm{t}=0 \times \mathrm{t}+\frac{1}{2} 5 \mathrm{t}^{2}\) \(\mathrm{t}=20 \mathrm{sec}\) Distance \(=\) speed \(\times\) time \(=50 \times 20=1000 \mathrm{~m}\)
Assam CEE-2017
Motion in One Dimensions
141258
A body is thrown vertically upwards. Which one of the following graphs correctly represent the velocity vs time?
141259
The displacement of a particle moving along \(x\) axis versus time is given in the figure below: The average velocity \(\left(\mathrm{V}_{\text {avg }}\right)\) of the particle in the first four seconds and the velocity \(V\) of it at \(t=4 s\) are
D Let us consider As \(\mathrm{t}=0\), particle is at a position \(5 \mathrm{~m}\), and at \(\mathrm{t}=4 \mathrm{~s}\) particle is at a position \(1 \mathrm{~m}\). Displacement of particle \((\mathrm{s})=1-5=-4 \mathrm{~m}\) time taken \((\mathrm{t})=4 \mathrm{~s}\) Average velocity \(\left(\mathrm{V}_{\text {avg }}\right)=\frac{\mathrm{s}}{\mathrm{t}}=\frac{-4}{4}=-1 \mathrm{~m} / \mathrm{s}\) At a slope of displacement time graph gives the velocity of particle at that instant. slope of the graph is zero at \(t=4 \mathrm{~s}\) then, the velocity is zero at \(t=4 \mathrm{~s}\) Hence, \(V_{a v}=-1 \mathrm{~m} / \mathrm{s}, V=0 \mathrm{~m} / \mathrm{s}\)
