362230
Velocity of a body is shown in the graph. The graph is a semicircle in shape. Find the displacement and distance in time from 0 \(s\) to 4 \(s\).
1 \(10\,\pi m\)
2 \(20\,\pi m\)
3 \(30\,\pi m\)
4 \(5\,\pi m\)
Explanation:
Displacement \( = \) distance (as the body is travelling in one direction though out time) \( = {\rm{Area}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {r_v}{r_t}}}{2}\) \( = \frac{{\pi \times 10 \times 2}}{2} = 10 \, \pi m\) Here \({r_v} - \) velocity radius, \({r_t} - \) time radius
PHXI03:MOTION IN A STRAIGHT LINE
362231
From the \(v-t\) graph shown, the ratio of distance to displacement in \(25 s\) of motion is
1 \(\dfrac{3}{5}\)
2 \(\dfrac{1}{2}\)
3 \(\dfrac{5}{3}\)
4 1
Explanation:
Area under the graph from \(t=0\) to \(t=20 \mathrm{sec}\), \(A_{1}=+\left[\dfrac{1}{2} \times 5 \times 10+5 \times 10+\dfrac{1}{2}(20+10)\right.\) \(\left.\times 5+\dfrac{1}{2} \times 5 \times 20\right]=+200 \mathrm{~m}\) Area under the graph from \(t=20\) to \(t=25 \mathrm{sec}\), \(A_{2}=-\dfrac{1}{2} \times 5 \times 20-50 m\left|A_{1}\right|+\left|A_{2}\right|\) So distance covered \( = \left| {{A_1}} \right| + \left| {{A_2}} \right|(200 + 50)m = 250\;m\) Displacement \(=A_{1}+A_{2}=200+(-50) m=150 m\) \(\therefore\) Ratio of distance to displacement is \(\dfrac{250}{150}=\dfrac{5}{3}\)
JEE - 2023
PHXI03:MOTION IN A STRAIGHT LINE
362232
Assertion : Displacement of a body may be zero when distance travelled by it is not zero. Reason : The displacement is the longest distance between initial and final position.
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:
The displacement is the shortest distance between initial and final position. When final position of a body coincides with its initial position, displacement is zero, but the distance travelled is not zero. So option (3) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362233
Velocity of a body starting from rest is shown in the diagram. Find the displacement and distance travelled by the body. From \(t = 0 \, s\) to \(8 \, s\)
1 \(12m,0m\)
2 \(24m,24m\)
3 \(24m,0m\)
4 \(16m,16m\)
Explanation:
As velocity is positive the body is moving in only one direction throughout the time. In this case displacement and distance are same. Distance \( = \) Displacement \( = \) Area of the graph \( = \left( {\frac{1}{2} \times 2 \times 4} \right) + (4 \times 4) + \left( {\frac{1}{2} \times 2 \times 4} \right)\) \( = 24\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362234
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
1 Neither can be zero
2 One may be zero
3 Both may be zero
4 One is \( + ve\), other is \( - ve\)
Explanation:
When location of a particle has changed,it must have covered some distance and undergone some displacement.
362230
Velocity of a body is shown in the graph. The graph is a semicircle in shape. Find the displacement and distance in time from 0 \(s\) to 4 \(s\).
1 \(10\,\pi m\)
2 \(20\,\pi m\)
3 \(30\,\pi m\)
4 \(5\,\pi m\)
Explanation:
Displacement \( = \) distance (as the body is travelling in one direction though out time) \( = {\rm{Area}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {r_v}{r_t}}}{2}\) \( = \frac{{\pi \times 10 \times 2}}{2} = 10 \, \pi m\) Here \({r_v} - \) velocity radius, \({r_t} - \) time radius
PHXI03:MOTION IN A STRAIGHT LINE
362231
From the \(v-t\) graph shown, the ratio of distance to displacement in \(25 s\) of motion is
1 \(\dfrac{3}{5}\)
2 \(\dfrac{1}{2}\)
3 \(\dfrac{5}{3}\)
4 1
Explanation:
Area under the graph from \(t=0\) to \(t=20 \mathrm{sec}\), \(A_{1}=+\left[\dfrac{1}{2} \times 5 \times 10+5 \times 10+\dfrac{1}{2}(20+10)\right.\) \(\left.\times 5+\dfrac{1}{2} \times 5 \times 20\right]=+200 \mathrm{~m}\) Area under the graph from \(t=20\) to \(t=25 \mathrm{sec}\), \(A_{2}=-\dfrac{1}{2} \times 5 \times 20-50 m\left|A_{1}\right|+\left|A_{2}\right|\) So distance covered \( = \left| {{A_1}} \right| + \left| {{A_2}} \right|(200 + 50)m = 250\;m\) Displacement \(=A_{1}+A_{2}=200+(-50) m=150 m\) \(\therefore\) Ratio of distance to displacement is \(\dfrac{250}{150}=\dfrac{5}{3}\)
JEE - 2023
PHXI03:MOTION IN A STRAIGHT LINE
362232
Assertion : Displacement of a body may be zero when distance travelled by it is not zero. Reason : The displacement is the longest distance between initial and final position.
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:
The displacement is the shortest distance between initial and final position. When final position of a body coincides with its initial position, displacement is zero, but the distance travelled is not zero. So option (3) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362233
Velocity of a body starting from rest is shown in the diagram. Find the displacement and distance travelled by the body. From \(t = 0 \, s\) to \(8 \, s\)
1 \(12m,0m\)
2 \(24m,24m\)
3 \(24m,0m\)
4 \(16m,16m\)
Explanation:
As velocity is positive the body is moving in only one direction throughout the time. In this case displacement and distance are same. Distance \( = \) Displacement \( = \) Area of the graph \( = \left( {\frac{1}{2} \times 2 \times 4} \right) + (4 \times 4) + \left( {\frac{1}{2} \times 2 \times 4} \right)\) \( = 24\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362234
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
1 Neither can be zero
2 One may be zero
3 Both may be zero
4 One is \( + ve\), other is \( - ve\)
Explanation:
When location of a particle has changed,it must have covered some distance and undergone some displacement.
362230
Velocity of a body is shown in the graph. The graph is a semicircle in shape. Find the displacement and distance in time from 0 \(s\) to 4 \(s\).
1 \(10\,\pi m\)
2 \(20\,\pi m\)
3 \(30\,\pi m\)
4 \(5\,\pi m\)
Explanation:
Displacement \( = \) distance (as the body is travelling in one direction though out time) \( = {\rm{Area}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {r_v}{r_t}}}{2}\) \( = \frac{{\pi \times 10 \times 2}}{2} = 10 \, \pi m\) Here \({r_v} - \) velocity radius, \({r_t} - \) time radius
PHXI03:MOTION IN A STRAIGHT LINE
362231
From the \(v-t\) graph shown, the ratio of distance to displacement in \(25 s\) of motion is
1 \(\dfrac{3}{5}\)
2 \(\dfrac{1}{2}\)
3 \(\dfrac{5}{3}\)
4 1
Explanation:
Area under the graph from \(t=0\) to \(t=20 \mathrm{sec}\), \(A_{1}=+\left[\dfrac{1}{2} \times 5 \times 10+5 \times 10+\dfrac{1}{2}(20+10)\right.\) \(\left.\times 5+\dfrac{1}{2} \times 5 \times 20\right]=+200 \mathrm{~m}\) Area under the graph from \(t=20\) to \(t=25 \mathrm{sec}\), \(A_{2}=-\dfrac{1}{2} \times 5 \times 20-50 m\left|A_{1}\right|+\left|A_{2}\right|\) So distance covered \( = \left| {{A_1}} \right| + \left| {{A_2}} \right|(200 + 50)m = 250\;m\) Displacement \(=A_{1}+A_{2}=200+(-50) m=150 m\) \(\therefore\) Ratio of distance to displacement is \(\dfrac{250}{150}=\dfrac{5}{3}\)
JEE - 2023
PHXI03:MOTION IN A STRAIGHT LINE
362232
Assertion : Displacement of a body may be zero when distance travelled by it is not zero. Reason : The displacement is the longest distance between initial and final position.
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:
The displacement is the shortest distance between initial and final position. When final position of a body coincides with its initial position, displacement is zero, but the distance travelled is not zero. So option (3) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362233
Velocity of a body starting from rest is shown in the diagram. Find the displacement and distance travelled by the body. From \(t = 0 \, s\) to \(8 \, s\)
1 \(12m,0m\)
2 \(24m,24m\)
3 \(24m,0m\)
4 \(16m,16m\)
Explanation:
As velocity is positive the body is moving in only one direction throughout the time. In this case displacement and distance are same. Distance \( = \) Displacement \( = \) Area of the graph \( = \left( {\frac{1}{2} \times 2 \times 4} \right) + (4 \times 4) + \left( {\frac{1}{2} \times 2 \times 4} \right)\) \( = 24\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362234
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
1 Neither can be zero
2 One may be zero
3 Both may be zero
4 One is \( + ve\), other is \( - ve\)
Explanation:
When location of a particle has changed,it must have covered some distance and undergone some displacement.
NEET Test Series from KOTA - 10 Papers In MS WORD
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PHXI03:MOTION IN A STRAIGHT LINE
362230
Velocity of a body is shown in the graph. The graph is a semicircle in shape. Find the displacement and distance in time from 0 \(s\) to 4 \(s\).
1 \(10\,\pi m\)
2 \(20\,\pi m\)
3 \(30\,\pi m\)
4 \(5\,\pi m\)
Explanation:
Displacement \( = \) distance (as the body is travelling in one direction though out time) \( = {\rm{Area}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {r_v}{r_t}}}{2}\) \( = \frac{{\pi \times 10 \times 2}}{2} = 10 \, \pi m\) Here \({r_v} - \) velocity radius, \({r_t} - \) time radius
PHXI03:MOTION IN A STRAIGHT LINE
362231
From the \(v-t\) graph shown, the ratio of distance to displacement in \(25 s\) of motion is
1 \(\dfrac{3}{5}\)
2 \(\dfrac{1}{2}\)
3 \(\dfrac{5}{3}\)
4 1
Explanation:
Area under the graph from \(t=0\) to \(t=20 \mathrm{sec}\), \(A_{1}=+\left[\dfrac{1}{2} \times 5 \times 10+5 \times 10+\dfrac{1}{2}(20+10)\right.\) \(\left.\times 5+\dfrac{1}{2} \times 5 \times 20\right]=+200 \mathrm{~m}\) Area under the graph from \(t=20\) to \(t=25 \mathrm{sec}\), \(A_{2}=-\dfrac{1}{2} \times 5 \times 20-50 m\left|A_{1}\right|+\left|A_{2}\right|\) So distance covered \( = \left| {{A_1}} \right| + \left| {{A_2}} \right|(200 + 50)m = 250\;m\) Displacement \(=A_{1}+A_{2}=200+(-50) m=150 m\) \(\therefore\) Ratio of distance to displacement is \(\dfrac{250}{150}=\dfrac{5}{3}\)
JEE - 2023
PHXI03:MOTION IN A STRAIGHT LINE
362232
Assertion : Displacement of a body may be zero when distance travelled by it is not zero. Reason : The displacement is the longest distance between initial and final position.
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:
The displacement is the shortest distance between initial and final position. When final position of a body coincides with its initial position, displacement is zero, but the distance travelled is not zero. So option (3) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362233
Velocity of a body starting from rest is shown in the diagram. Find the displacement and distance travelled by the body. From \(t = 0 \, s\) to \(8 \, s\)
1 \(12m,0m\)
2 \(24m,24m\)
3 \(24m,0m\)
4 \(16m,16m\)
Explanation:
As velocity is positive the body is moving in only one direction throughout the time. In this case displacement and distance are same. Distance \( = \) Displacement \( = \) Area of the graph \( = \left( {\frac{1}{2} \times 2 \times 4} \right) + (4 \times 4) + \left( {\frac{1}{2} \times 2 \times 4} \right)\) \( = 24\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362234
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
1 Neither can be zero
2 One may be zero
3 Both may be zero
4 One is \( + ve\), other is \( - ve\)
Explanation:
When location of a particle has changed,it must have covered some distance and undergone some displacement.
362230
Velocity of a body is shown in the graph. The graph is a semicircle in shape. Find the displacement and distance in time from 0 \(s\) to 4 \(s\).
1 \(10\,\pi m\)
2 \(20\,\pi m\)
3 \(30\,\pi m\)
4 \(5\,\pi m\)
Explanation:
Displacement \( = \) distance (as the body is travelling in one direction though out time) \( = {\rm{Area}} = \frac{{\pi {r^2}}}{2} = \frac{{\pi {r_v}{r_t}}}{2}\) \( = \frac{{\pi \times 10 \times 2}}{2} = 10 \, \pi m\) Here \({r_v} - \) velocity radius, \({r_t} - \) time radius
PHXI03:MOTION IN A STRAIGHT LINE
362231
From the \(v-t\) graph shown, the ratio of distance to displacement in \(25 s\) of motion is
1 \(\dfrac{3}{5}\)
2 \(\dfrac{1}{2}\)
3 \(\dfrac{5}{3}\)
4 1
Explanation:
Area under the graph from \(t=0\) to \(t=20 \mathrm{sec}\), \(A_{1}=+\left[\dfrac{1}{2} \times 5 \times 10+5 \times 10+\dfrac{1}{2}(20+10)\right.\) \(\left.\times 5+\dfrac{1}{2} \times 5 \times 20\right]=+200 \mathrm{~m}\) Area under the graph from \(t=20\) to \(t=25 \mathrm{sec}\), \(A_{2}=-\dfrac{1}{2} \times 5 \times 20-50 m\left|A_{1}\right|+\left|A_{2}\right|\) So distance covered \( = \left| {{A_1}} \right| + \left| {{A_2}} \right|(200 + 50)m = 250\;m\) Displacement \(=A_{1}+A_{2}=200+(-50) m=150 m\) \(\therefore\) Ratio of distance to displacement is \(\dfrac{250}{150}=\dfrac{5}{3}\)
JEE - 2023
PHXI03:MOTION IN A STRAIGHT LINE
362232
Assertion : Displacement of a body may be zero when distance travelled by it is not zero. Reason : The displacement is the longest distance between initial and final position.
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:
The displacement is the shortest distance between initial and final position. When final position of a body coincides with its initial position, displacement is zero, but the distance travelled is not zero. So option (3) is correct.
PHXI03:MOTION IN A STRAIGHT LINE
362233
Velocity of a body starting from rest is shown in the diagram. Find the displacement and distance travelled by the body. From \(t = 0 \, s\) to \(8 \, s\)
1 \(12m,0m\)
2 \(24m,24m\)
3 \(24m,0m\)
4 \(16m,16m\)
Explanation:
As velocity is positive the body is moving in only one direction throughout the time. In this case displacement and distance are same. Distance \( = \) Displacement \( = \) Area of the graph \( = \left( {\frac{1}{2} \times 2 \times 4} \right) + (4 \times 4) + \left( {\frac{1}{2} \times 2 \times 4} \right)\) \( = 24\,m\)
PHXI03:MOTION IN A STRAIGHT LINE
362234
The location of a particle has changed. What can we say about the displacement and the distance covered by the particle?
1 Neither can be zero
2 One may be zero
3 Both may be zero
4 One is \( + ve\), other is \( - ve\)
Explanation:
When location of a particle has changed,it must have covered some distance and undergone some displacement.
