141472
A body covers \(2 / 5^{\text {th }}\) of the total distance with speed \(v_{1}\) and \(3 / 5^{\text {th }}\) with \(v_{2}\). The average speed of the body will be-
A Let the total distance is \(x\), then we have \(\mathrm{t}_{1}=\frac{2}{5} \frac{\mathrm{x}}{\mathrm{v}_{1}} \text { and } \mathrm{t}_{2}=\frac{3}{5} \frac{\mathrm{x}}{\mathrm{v}_{2}}\) Since average speed, \(\quad \mathbf{v}_{\mathrm{av}}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\mathrm{x}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\) \(=\frac{\mathrm{x}}{\frac{2 \mathrm{x}}{5 \mathrm{v}_{1}}+\frac{3 \mathrm{x}}{5 \mathrm{v}_{2}}}=\frac{\mathrm{x}}{\frac{10 \mathrm{xv}_{2}+15 \mathrm{xv}_{1}}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(=\frac{\mathrm{x}}{\frac{5 \mathrm{x}\left(2 \mathrm{v}_{2}+3 \mathrm{v}_{1}\right)}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(\therefore \quad=\mathrm{v}_{\mathrm{av}}=\frac{5 \mathrm{v}_{1} \mathrm{v}_{2}}{3 \mathrm{v}_{1}+2 \mathrm{v}_{2}}\)
BCECE-2018
Motion in One Dimensions
141473
The velocity-time graph of two bodies \(A\) and \(B\) are shown below. Then, the ratio of their acceleration \(a_{A} / a_{b}\) will be-
1 \(\sqrt{3}: 1\)
2 \(1: \sqrt{3}\)
3 \(3: 1\)
4 \(1: 3\)
Explanation:
D Acceleration = Slope of velocity - time graph Thus, \(\quad \frac{a_{A}}{a_{B}}=\frac{\text { Slope of } v-t \text { graph for } A}{\text { Slope of } v-t \text { graph for } B}\) \(=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\) \(\frac{a_{A}}{a_{B}}=\frac{1 / \sqrt{3}}{\sqrt{3}}=\frac{1}{3} \Rightarrow a_{A}: a_{B}=1: 3\)
BCECE-2018
Motion in One Dimensions
141474
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_{1}\) in the first 10 seconds and distance \(s_{2}\) in the next 10 seconds, then
1 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
2 \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Explanation:
C Let a be the constant acceleration of the particle. From the second equation of motion- \(\quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\text { or } \quad \mathrm{s}_{1}=0+\frac{1}{2} \times \mathrm{a} \times(10)^{2}=50 \mathrm{a}\) \(\mathrm{S}_{2}=(\text { Total distance covered in } 20 \mathrm{sec})-(\text { total distance }\) \(\text { covered in first } 10 \text { second })\) \(\mathrm{S}_{2}=\mathrm{S}_{20}-\mathrm{S}_{10}\) \(\text { and } \quad \mathrm{s}_{2}=\left[0+\frac{1}{2} \times \mathrm{a} \times(20)^{2}\right]-50 \mathrm{a}=150 \mathrm{a}\) \(\therefore \quad \mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
AIIMS-27.05.2018(M)
Motion in One Dimensions
141475
A bus is moving with a velocity of \(10 \mathrm{~ms}^{-1}\) on a straight road. A scootorist wishes to overtake the bus in one minute. If the bus is at a distance of \(1.2 \mathrm{~km}\) ahead, then the velocity with which he has to chase the bus is
1 \(20 \mathrm{~ms}^{-1}\)
2 \(25 \mathrm{~ms}^{-1}\)
3 \(60 \mathrm{~ms}^{-1}\)
4 \(30 \mathrm{~ms}^{-1}\)
Explanation:
D Given, \(\mathrm{t}=60 \mathrm{sec}, \text { Distance }=1.2 \mathrm{~km}=1200 \mathrm{~m}\) \(\mathrm{v}_{\text {bus }}=10 \mathrm{~m} / \mathrm{s} \) The relative speed with respect to the bus is - \(\mathbf{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{1200}{60}=20 \mathrm{~m} / \mathrm{s}\) The scooter and the bus are moving along the same direction, so- \(\text { speed } (\mathrm{v})=\mathrm{v}_{\text {scooter }}-\mathrm{v} \text { bus }\) \(20=\mathrm{v}_{\text {scooter }}-10\) \(\mathrm{v}_{\text {scooter }}=30 \mathrm{~m} / \mathrm{s}\)
AIIMS-26.05.2018(M)
Motion in One Dimensions
141476
The motion of a particle along a straight line is described by equation: \(x=8+12 t-t^{3}\) where \(x\) is in meter and \(t\) is second. The retardation of the particle when its velocity becomes zero is:
1 \(24 \mathrm{~ms}^{-2}\)
2 zero
3 \(6 \mathrm{~ms}^{-2}\)
4 \(12 \mathrm{~ms}^{-2}\)
Explanation:
D Given, \(\mathrm{x}=8+12 \mathrm{t}-\mathrm{t}^{3}\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=12-3 \mathrm{t}^{2}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6 \mathrm{t}\) Putting \(\mathrm{v}=0\) in equ. (i) \(3 \mathrm{t}^{2}=12\) \(\mathrm{t}=2 \mathrm{sec}\) From equ. (ii) \(\quad \mathrm{a}=-6 \times 2=-12 \mathrm{~m} / \mathrm{s}^{2}\) Retardation \(=12 \mathrm{~m} / \mathrm{s}^{2}\)
141472
A body covers \(2 / 5^{\text {th }}\) of the total distance with speed \(v_{1}\) and \(3 / 5^{\text {th }}\) with \(v_{2}\). The average speed of the body will be-
A Let the total distance is \(x\), then we have \(\mathrm{t}_{1}=\frac{2}{5} \frac{\mathrm{x}}{\mathrm{v}_{1}} \text { and } \mathrm{t}_{2}=\frac{3}{5} \frac{\mathrm{x}}{\mathrm{v}_{2}}\) Since average speed, \(\quad \mathbf{v}_{\mathrm{av}}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\mathrm{x}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\) \(=\frac{\mathrm{x}}{\frac{2 \mathrm{x}}{5 \mathrm{v}_{1}}+\frac{3 \mathrm{x}}{5 \mathrm{v}_{2}}}=\frac{\mathrm{x}}{\frac{10 \mathrm{xv}_{2}+15 \mathrm{xv}_{1}}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(=\frac{\mathrm{x}}{\frac{5 \mathrm{x}\left(2 \mathrm{v}_{2}+3 \mathrm{v}_{1}\right)}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(\therefore \quad=\mathrm{v}_{\mathrm{av}}=\frac{5 \mathrm{v}_{1} \mathrm{v}_{2}}{3 \mathrm{v}_{1}+2 \mathrm{v}_{2}}\)
BCECE-2018
Motion in One Dimensions
141473
The velocity-time graph of two bodies \(A\) and \(B\) are shown below. Then, the ratio of their acceleration \(a_{A} / a_{b}\) will be-
1 \(\sqrt{3}: 1\)
2 \(1: \sqrt{3}\)
3 \(3: 1\)
4 \(1: 3\)
Explanation:
D Acceleration = Slope of velocity - time graph Thus, \(\quad \frac{a_{A}}{a_{B}}=\frac{\text { Slope of } v-t \text { graph for } A}{\text { Slope of } v-t \text { graph for } B}\) \(=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\) \(\frac{a_{A}}{a_{B}}=\frac{1 / \sqrt{3}}{\sqrt{3}}=\frac{1}{3} \Rightarrow a_{A}: a_{B}=1: 3\)
BCECE-2018
Motion in One Dimensions
141474
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_{1}\) in the first 10 seconds and distance \(s_{2}\) in the next 10 seconds, then
1 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
2 \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Explanation:
C Let a be the constant acceleration of the particle. From the second equation of motion- \(\quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\text { or } \quad \mathrm{s}_{1}=0+\frac{1}{2} \times \mathrm{a} \times(10)^{2}=50 \mathrm{a}\) \(\mathrm{S}_{2}=(\text { Total distance covered in } 20 \mathrm{sec})-(\text { total distance }\) \(\text { covered in first } 10 \text { second })\) \(\mathrm{S}_{2}=\mathrm{S}_{20}-\mathrm{S}_{10}\) \(\text { and } \quad \mathrm{s}_{2}=\left[0+\frac{1}{2} \times \mathrm{a} \times(20)^{2}\right]-50 \mathrm{a}=150 \mathrm{a}\) \(\therefore \quad \mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
AIIMS-27.05.2018(M)
Motion in One Dimensions
141475
A bus is moving with a velocity of \(10 \mathrm{~ms}^{-1}\) on a straight road. A scootorist wishes to overtake the bus in one minute. If the bus is at a distance of \(1.2 \mathrm{~km}\) ahead, then the velocity with which he has to chase the bus is
1 \(20 \mathrm{~ms}^{-1}\)
2 \(25 \mathrm{~ms}^{-1}\)
3 \(60 \mathrm{~ms}^{-1}\)
4 \(30 \mathrm{~ms}^{-1}\)
Explanation:
D Given, \(\mathrm{t}=60 \mathrm{sec}, \text { Distance }=1.2 \mathrm{~km}=1200 \mathrm{~m}\) \(\mathrm{v}_{\text {bus }}=10 \mathrm{~m} / \mathrm{s} \) The relative speed with respect to the bus is - \(\mathbf{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{1200}{60}=20 \mathrm{~m} / \mathrm{s}\) The scooter and the bus are moving along the same direction, so- \(\text { speed } (\mathrm{v})=\mathrm{v}_{\text {scooter }}-\mathrm{v} \text { bus }\) \(20=\mathrm{v}_{\text {scooter }}-10\) \(\mathrm{v}_{\text {scooter }}=30 \mathrm{~m} / \mathrm{s}\)
AIIMS-26.05.2018(M)
Motion in One Dimensions
141476
The motion of a particle along a straight line is described by equation: \(x=8+12 t-t^{3}\) where \(x\) is in meter and \(t\) is second. The retardation of the particle when its velocity becomes zero is:
1 \(24 \mathrm{~ms}^{-2}\)
2 zero
3 \(6 \mathrm{~ms}^{-2}\)
4 \(12 \mathrm{~ms}^{-2}\)
Explanation:
D Given, \(\mathrm{x}=8+12 \mathrm{t}-\mathrm{t}^{3}\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=12-3 \mathrm{t}^{2}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6 \mathrm{t}\) Putting \(\mathrm{v}=0\) in equ. (i) \(3 \mathrm{t}^{2}=12\) \(\mathrm{t}=2 \mathrm{sec}\) From equ. (ii) \(\quad \mathrm{a}=-6 \times 2=-12 \mathrm{~m} / \mathrm{s}^{2}\) Retardation \(=12 \mathrm{~m} / \mathrm{s}^{2}\)
141472
A body covers \(2 / 5^{\text {th }}\) of the total distance with speed \(v_{1}\) and \(3 / 5^{\text {th }}\) with \(v_{2}\). The average speed of the body will be-
A Let the total distance is \(x\), then we have \(\mathrm{t}_{1}=\frac{2}{5} \frac{\mathrm{x}}{\mathrm{v}_{1}} \text { and } \mathrm{t}_{2}=\frac{3}{5} \frac{\mathrm{x}}{\mathrm{v}_{2}}\) Since average speed, \(\quad \mathbf{v}_{\mathrm{av}}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\mathrm{x}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\) \(=\frac{\mathrm{x}}{\frac{2 \mathrm{x}}{5 \mathrm{v}_{1}}+\frac{3 \mathrm{x}}{5 \mathrm{v}_{2}}}=\frac{\mathrm{x}}{\frac{10 \mathrm{xv}_{2}+15 \mathrm{xv}_{1}}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(=\frac{\mathrm{x}}{\frac{5 \mathrm{x}\left(2 \mathrm{v}_{2}+3 \mathrm{v}_{1}\right)}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(\therefore \quad=\mathrm{v}_{\mathrm{av}}=\frac{5 \mathrm{v}_{1} \mathrm{v}_{2}}{3 \mathrm{v}_{1}+2 \mathrm{v}_{2}}\)
BCECE-2018
Motion in One Dimensions
141473
The velocity-time graph of two bodies \(A\) and \(B\) are shown below. Then, the ratio of their acceleration \(a_{A} / a_{b}\) will be-
1 \(\sqrt{3}: 1\)
2 \(1: \sqrt{3}\)
3 \(3: 1\)
4 \(1: 3\)
Explanation:
D Acceleration = Slope of velocity - time graph Thus, \(\quad \frac{a_{A}}{a_{B}}=\frac{\text { Slope of } v-t \text { graph for } A}{\text { Slope of } v-t \text { graph for } B}\) \(=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\) \(\frac{a_{A}}{a_{B}}=\frac{1 / \sqrt{3}}{\sqrt{3}}=\frac{1}{3} \Rightarrow a_{A}: a_{B}=1: 3\)
BCECE-2018
Motion in One Dimensions
141474
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_{1}\) in the first 10 seconds and distance \(s_{2}\) in the next 10 seconds, then
1 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
2 \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Explanation:
C Let a be the constant acceleration of the particle. From the second equation of motion- \(\quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\text { or } \quad \mathrm{s}_{1}=0+\frac{1}{2} \times \mathrm{a} \times(10)^{2}=50 \mathrm{a}\) \(\mathrm{S}_{2}=(\text { Total distance covered in } 20 \mathrm{sec})-(\text { total distance }\) \(\text { covered in first } 10 \text { second })\) \(\mathrm{S}_{2}=\mathrm{S}_{20}-\mathrm{S}_{10}\) \(\text { and } \quad \mathrm{s}_{2}=\left[0+\frac{1}{2} \times \mathrm{a} \times(20)^{2}\right]-50 \mathrm{a}=150 \mathrm{a}\) \(\therefore \quad \mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
AIIMS-27.05.2018(M)
Motion in One Dimensions
141475
A bus is moving with a velocity of \(10 \mathrm{~ms}^{-1}\) on a straight road. A scootorist wishes to overtake the bus in one minute. If the bus is at a distance of \(1.2 \mathrm{~km}\) ahead, then the velocity with which he has to chase the bus is
1 \(20 \mathrm{~ms}^{-1}\)
2 \(25 \mathrm{~ms}^{-1}\)
3 \(60 \mathrm{~ms}^{-1}\)
4 \(30 \mathrm{~ms}^{-1}\)
Explanation:
D Given, \(\mathrm{t}=60 \mathrm{sec}, \text { Distance }=1.2 \mathrm{~km}=1200 \mathrm{~m}\) \(\mathrm{v}_{\text {bus }}=10 \mathrm{~m} / \mathrm{s} \) The relative speed with respect to the bus is - \(\mathbf{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{1200}{60}=20 \mathrm{~m} / \mathrm{s}\) The scooter and the bus are moving along the same direction, so- \(\text { speed } (\mathrm{v})=\mathrm{v}_{\text {scooter }}-\mathrm{v} \text { bus }\) \(20=\mathrm{v}_{\text {scooter }}-10\) \(\mathrm{v}_{\text {scooter }}=30 \mathrm{~m} / \mathrm{s}\)
AIIMS-26.05.2018(M)
Motion in One Dimensions
141476
The motion of a particle along a straight line is described by equation: \(x=8+12 t-t^{3}\) where \(x\) is in meter and \(t\) is second. The retardation of the particle when its velocity becomes zero is:
1 \(24 \mathrm{~ms}^{-2}\)
2 zero
3 \(6 \mathrm{~ms}^{-2}\)
4 \(12 \mathrm{~ms}^{-2}\)
Explanation:
D Given, \(\mathrm{x}=8+12 \mathrm{t}-\mathrm{t}^{3}\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=12-3 \mathrm{t}^{2}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6 \mathrm{t}\) Putting \(\mathrm{v}=0\) in equ. (i) \(3 \mathrm{t}^{2}=12\) \(\mathrm{t}=2 \mathrm{sec}\) From equ. (ii) \(\quad \mathrm{a}=-6 \times 2=-12 \mathrm{~m} / \mathrm{s}^{2}\) Retardation \(=12 \mathrm{~m} / \mathrm{s}^{2}\)
141472
A body covers \(2 / 5^{\text {th }}\) of the total distance with speed \(v_{1}\) and \(3 / 5^{\text {th }}\) with \(v_{2}\). The average speed of the body will be-
A Let the total distance is \(x\), then we have \(\mathrm{t}_{1}=\frac{2}{5} \frac{\mathrm{x}}{\mathrm{v}_{1}} \text { and } \mathrm{t}_{2}=\frac{3}{5} \frac{\mathrm{x}}{\mathrm{v}_{2}}\) Since average speed, \(\quad \mathbf{v}_{\mathrm{av}}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\mathrm{x}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\) \(=\frac{\mathrm{x}}{\frac{2 \mathrm{x}}{5 \mathrm{v}_{1}}+\frac{3 \mathrm{x}}{5 \mathrm{v}_{2}}}=\frac{\mathrm{x}}{\frac{10 \mathrm{xv}_{2}+15 \mathrm{xv}_{1}}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(=\frac{\mathrm{x}}{\frac{5 \mathrm{x}\left(2 \mathrm{v}_{2}+3 \mathrm{v}_{1}\right)}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(\therefore \quad=\mathrm{v}_{\mathrm{av}}=\frac{5 \mathrm{v}_{1} \mathrm{v}_{2}}{3 \mathrm{v}_{1}+2 \mathrm{v}_{2}}\)
BCECE-2018
Motion in One Dimensions
141473
The velocity-time graph of two bodies \(A\) and \(B\) are shown below. Then, the ratio of their acceleration \(a_{A} / a_{b}\) will be-
1 \(\sqrt{3}: 1\)
2 \(1: \sqrt{3}\)
3 \(3: 1\)
4 \(1: 3\)
Explanation:
D Acceleration = Slope of velocity - time graph Thus, \(\quad \frac{a_{A}}{a_{B}}=\frac{\text { Slope of } v-t \text { graph for } A}{\text { Slope of } v-t \text { graph for } B}\) \(=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\) \(\frac{a_{A}}{a_{B}}=\frac{1 / \sqrt{3}}{\sqrt{3}}=\frac{1}{3} \Rightarrow a_{A}: a_{B}=1: 3\)
BCECE-2018
Motion in One Dimensions
141474
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_{1}\) in the first 10 seconds and distance \(s_{2}\) in the next 10 seconds, then
1 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
2 \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Explanation:
C Let a be the constant acceleration of the particle. From the second equation of motion- \(\quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\text { or } \quad \mathrm{s}_{1}=0+\frac{1}{2} \times \mathrm{a} \times(10)^{2}=50 \mathrm{a}\) \(\mathrm{S}_{2}=(\text { Total distance covered in } 20 \mathrm{sec})-(\text { total distance }\) \(\text { covered in first } 10 \text { second })\) \(\mathrm{S}_{2}=\mathrm{S}_{20}-\mathrm{S}_{10}\) \(\text { and } \quad \mathrm{s}_{2}=\left[0+\frac{1}{2} \times \mathrm{a} \times(20)^{2}\right]-50 \mathrm{a}=150 \mathrm{a}\) \(\therefore \quad \mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
AIIMS-27.05.2018(M)
Motion in One Dimensions
141475
A bus is moving with a velocity of \(10 \mathrm{~ms}^{-1}\) on a straight road. A scootorist wishes to overtake the bus in one minute. If the bus is at a distance of \(1.2 \mathrm{~km}\) ahead, then the velocity with which he has to chase the bus is
1 \(20 \mathrm{~ms}^{-1}\)
2 \(25 \mathrm{~ms}^{-1}\)
3 \(60 \mathrm{~ms}^{-1}\)
4 \(30 \mathrm{~ms}^{-1}\)
Explanation:
D Given, \(\mathrm{t}=60 \mathrm{sec}, \text { Distance }=1.2 \mathrm{~km}=1200 \mathrm{~m}\) \(\mathrm{v}_{\text {bus }}=10 \mathrm{~m} / \mathrm{s} \) The relative speed with respect to the bus is - \(\mathbf{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{1200}{60}=20 \mathrm{~m} / \mathrm{s}\) The scooter and the bus are moving along the same direction, so- \(\text { speed } (\mathrm{v})=\mathrm{v}_{\text {scooter }}-\mathrm{v} \text { bus }\) \(20=\mathrm{v}_{\text {scooter }}-10\) \(\mathrm{v}_{\text {scooter }}=30 \mathrm{~m} / \mathrm{s}\)
AIIMS-26.05.2018(M)
Motion in One Dimensions
141476
The motion of a particle along a straight line is described by equation: \(x=8+12 t-t^{3}\) where \(x\) is in meter and \(t\) is second. The retardation of the particle when its velocity becomes zero is:
1 \(24 \mathrm{~ms}^{-2}\)
2 zero
3 \(6 \mathrm{~ms}^{-2}\)
4 \(12 \mathrm{~ms}^{-2}\)
Explanation:
D Given, \(\mathrm{x}=8+12 \mathrm{t}-\mathrm{t}^{3}\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=12-3 \mathrm{t}^{2}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6 \mathrm{t}\) Putting \(\mathrm{v}=0\) in equ. (i) \(3 \mathrm{t}^{2}=12\) \(\mathrm{t}=2 \mathrm{sec}\) From equ. (ii) \(\quad \mathrm{a}=-6 \times 2=-12 \mathrm{~m} / \mathrm{s}^{2}\) Retardation \(=12 \mathrm{~m} / \mathrm{s}^{2}\)
141472
A body covers \(2 / 5^{\text {th }}\) of the total distance with speed \(v_{1}\) and \(3 / 5^{\text {th }}\) with \(v_{2}\). The average speed of the body will be-
A Let the total distance is \(x\), then we have \(\mathrm{t}_{1}=\frac{2}{5} \frac{\mathrm{x}}{\mathrm{v}_{1}} \text { and } \mathrm{t}_{2}=\frac{3}{5} \frac{\mathrm{x}}{\mathrm{v}_{2}}\) Since average speed, \(\quad \mathbf{v}_{\mathrm{av}}=\frac{\text { Total distance }}{\text { Total time }}=\frac{\mathrm{x}}{\mathrm{t}_{1}+\mathrm{t}_{2}}\) \(=\frac{\mathrm{x}}{\frac{2 \mathrm{x}}{5 \mathrm{v}_{1}}+\frac{3 \mathrm{x}}{5 \mathrm{v}_{2}}}=\frac{\mathrm{x}}{\frac{10 \mathrm{xv}_{2}+15 \mathrm{xv}_{1}}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(=\frac{\mathrm{x}}{\frac{5 \mathrm{x}\left(2 \mathrm{v}_{2}+3 \mathrm{v}_{1}\right)}{25 \mathrm{v}_{1} \mathrm{v}_{2}}}\) \(\therefore \quad=\mathrm{v}_{\mathrm{av}}=\frac{5 \mathrm{v}_{1} \mathrm{v}_{2}}{3 \mathrm{v}_{1}+2 \mathrm{v}_{2}}\)
BCECE-2018
Motion in One Dimensions
141473
The velocity-time graph of two bodies \(A\) and \(B\) are shown below. Then, the ratio of their acceleration \(a_{A} / a_{b}\) will be-
1 \(\sqrt{3}: 1\)
2 \(1: \sqrt{3}\)
3 \(3: 1\)
4 \(1: 3\)
Explanation:
D Acceleration = Slope of velocity - time graph Thus, \(\quad \frac{a_{A}}{a_{B}}=\frac{\text { Slope of } v-t \text { graph for } A}{\text { Slope of } v-t \text { graph for } B}\) \(=\frac{\tan 30^{\circ}}{\tan 60^{\circ}}\) \(\frac{a_{A}}{a_{B}}=\frac{1 / \sqrt{3}}{\sqrt{3}}=\frac{1}{3} \Rightarrow a_{A}: a_{B}=1: 3\)
BCECE-2018
Motion in One Dimensions
141474
A particle experiences constant acceleration for 20 seconds after starting from rest. If it travels a distance \(s_{1}\) in the first 10 seconds and distance \(s_{2}\) in the next 10 seconds, then
1 \(\mathrm{s}_{2}=\mathrm{s}_{1}\)
2 \(\mathrm{s}_{2}=2 \mathrm{~s}_{1}\)
3 \(\mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
4 \(\mathrm{s}_{2}=4 \mathrm{~s}_{1}\)
Explanation:
C Let a be the constant acceleration of the particle. From the second equation of motion- \(\quad \mathrm{s}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^{2}\) \(\text { or } \quad \mathrm{s}_{1}=0+\frac{1}{2} \times \mathrm{a} \times(10)^{2}=50 \mathrm{a}\) \(\mathrm{S}_{2}=(\text { Total distance covered in } 20 \mathrm{sec})-(\text { total distance }\) \(\text { covered in first } 10 \text { second })\) \(\mathrm{S}_{2}=\mathrm{S}_{20}-\mathrm{S}_{10}\) \(\text { and } \quad \mathrm{s}_{2}=\left[0+\frac{1}{2} \times \mathrm{a} \times(20)^{2}\right]-50 \mathrm{a}=150 \mathrm{a}\) \(\therefore \quad \mathrm{s}_{2}=3 \mathrm{~s}_{1}\)
AIIMS-27.05.2018(M)
Motion in One Dimensions
141475
A bus is moving with a velocity of \(10 \mathrm{~ms}^{-1}\) on a straight road. A scootorist wishes to overtake the bus in one minute. If the bus is at a distance of \(1.2 \mathrm{~km}\) ahead, then the velocity with which he has to chase the bus is
1 \(20 \mathrm{~ms}^{-1}\)
2 \(25 \mathrm{~ms}^{-1}\)
3 \(60 \mathrm{~ms}^{-1}\)
4 \(30 \mathrm{~ms}^{-1}\)
Explanation:
D Given, \(\mathrm{t}=60 \mathrm{sec}, \text { Distance }=1.2 \mathrm{~km}=1200 \mathrm{~m}\) \(\mathrm{v}_{\text {bus }}=10 \mathrm{~m} / \mathrm{s} \) The relative speed with respect to the bus is - \(\mathbf{v}=\frac{\mathrm{d}}{\mathrm{t}}=\frac{1200}{60}=20 \mathrm{~m} / \mathrm{s}\) The scooter and the bus are moving along the same direction, so- \(\text { speed } (\mathrm{v})=\mathrm{v}_{\text {scooter }}-\mathrm{v} \text { bus }\) \(20=\mathrm{v}_{\text {scooter }}-10\) \(\mathrm{v}_{\text {scooter }}=30 \mathrm{~m} / \mathrm{s}\)
AIIMS-26.05.2018(M)
Motion in One Dimensions
141476
The motion of a particle along a straight line is described by equation: \(x=8+12 t-t^{3}\) where \(x\) is in meter and \(t\) is second. The retardation of the particle when its velocity becomes zero is:
1 \(24 \mathrm{~ms}^{-2}\)
2 zero
3 \(6 \mathrm{~ms}^{-2}\)
4 \(12 \mathrm{~ms}^{-2}\)
Explanation:
D Given, \(\mathrm{x}=8+12 \mathrm{t}-\mathrm{t}^{3}\) \(\mathrm{v}=\frac{\mathrm{dx}}{\mathrm{dt}}=12-3 \mathrm{t}^{2}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-6 \mathrm{t}\) Putting \(\mathrm{v}=0\) in equ. (i) \(3 \mathrm{t}^{2}=12\) \(\mathrm{t}=2 \mathrm{sec}\) From equ. (ii) \(\quad \mathrm{a}=-6 \times 2=-12 \mathrm{~m} / \mathrm{s}^{2}\) Retardation \(=12 \mathrm{~m} / \mathrm{s}^{2}\)
