141418
The stopping distance of a moving vehicle is proportional to the
1 initial velocity
2 cube of the initial velocity
3 square of the initial velocity
4 cube root of the initial velocity
5 square root of the initial velocity
Explanation:
C The stopping distance of a moving vehicle is propotional to the square of the initial velocity. Let, Initial velocity \(=\mathrm{u}\) Final velocity \((v)=0\) Using third equation of motion \(v^{2}=u^{2}+2 a s\) \((0)^{2}=u^{2}+2 a s\) \(-u^{2}=-2 a s \quad \quad[\because a=\text { acceleration }\) \(\left.s=\frac{u^{2}}{2 a} \quad s=\text { stopping, side distance }\right]\) \(s \propto u^{2}\)
Kerala CEE 2021
Motion in One Dimensions
141419
The displacement of a particle starting from rest a \(t=0\) is given by \(s=9 t^2-2 t^3\). The time in seconds at which the particle will attain zero velocity is
1 \(8 \mathrm{~s}\)
2 \(6 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(3 \mathrm{~s}\)
Explanation:
D Given data, \(s=9 t^2-2 t^3\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=18 \mathrm{t}-6 \mathrm{t}^2\) \(\mathrm{v} =18 \mathrm{t}-6 \mathrm{t}^2\) \(0 =6 \mathrm{t}(3-\mathrm{t})\) \(0=6 \mathrm{t}(3-\mathrm{t}) \quad\{\because \mathrm{v}=0\}\) \(\mathrm{t} =0 \text { or } 3 \mathrm{sec}\) So, the particle will attain zero velocity at the time of 3 seconds. So, option (d) is correct.
AP EAMCET-20.08.2021
Motion in One Dimensions
141420
The slope of velocity - time graph for motion with uniform velocity is equal to:
1 Initial velocity
2 Final velocity
3 Zero
4 Constant velocity
Explanation:
C When velocity is uniform, velocity time graph is straight line parallel to the time axis. So, the slope is zero. For constant velocity acceleration, \(\mathrm{a}=0\) \(\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\) \(\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\) So, option (c) is correct
AP EAMCET-24.08.2021
Motion in One Dimensions
141421
The displacement of a particle moving along \(x\) axis is given by \(x=8 t^{2}+18 t\). The average acceleration during the interval \(t_{1}=2\) and \(t_{2}=\) \(4 \mathrm{~s}\) is
D According to equation, The displacement of the particle is given by, \(x =8 t^{2}+18 t\) \(\therefore v =\frac{d x}{d t}=16 t+18\) \(v =16 t+18\) at \(\mathrm{t}_{1}=2 \mathrm{sec} \Rightarrow \mathrm{v}_{1}=16 \times 2+18=32+18=50 \mathrm{~m} / \mathrm{s}\) at \(\mathrm{t}_{2}=4 \mathrm{sec} \Rightarrow \mathrm{v}_{2}=16 \times 4+18=64+18=82 \mathrm{~m} / \mathrm{s}\) average acceleration, \(\mathrm{a}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{t}}=\frac{82-50}{2}=\frac{32}{2}=16\) \(\mathrm{a}=16 \mathrm{~ms}^{-2}\) So, option (d) is correct
AP EAMCET-03.09.2021
Motion in One Dimensions
141422
The velocity-time graph of a moving object is shown in the figure. Find its maximum acceleration?
1 \(3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(6 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(4 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
B Acceleration in the \(1^{\text {st }}\) part. \(\mathrm{a}_{1}=\frac{20-0}{30-0}=0.66 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(2^{\text {nd }}\) part. \(\mathrm{a}_{2}=\frac{80-20}{40-30}=6 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(3^{\text {rd }}\) part. \(\mathrm{a}_{3}=\frac{0-80}{80-40}=-2 \mathrm{~m} / \mathrm{s}^{2}\) Max. acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\)
141418
The stopping distance of a moving vehicle is proportional to the
1 initial velocity
2 cube of the initial velocity
3 square of the initial velocity
4 cube root of the initial velocity
5 square root of the initial velocity
Explanation:
C The stopping distance of a moving vehicle is propotional to the square of the initial velocity. Let, Initial velocity \(=\mathrm{u}\) Final velocity \((v)=0\) Using third equation of motion \(v^{2}=u^{2}+2 a s\) \((0)^{2}=u^{2}+2 a s\) \(-u^{2}=-2 a s \quad \quad[\because a=\text { acceleration }\) \(\left.s=\frac{u^{2}}{2 a} \quad s=\text { stopping, side distance }\right]\) \(s \propto u^{2}\)
Kerala CEE 2021
Motion in One Dimensions
141419
The displacement of a particle starting from rest a \(t=0\) is given by \(s=9 t^2-2 t^3\). The time in seconds at which the particle will attain zero velocity is
1 \(8 \mathrm{~s}\)
2 \(6 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(3 \mathrm{~s}\)
Explanation:
D Given data, \(s=9 t^2-2 t^3\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=18 \mathrm{t}-6 \mathrm{t}^2\) \(\mathrm{v} =18 \mathrm{t}-6 \mathrm{t}^2\) \(0 =6 \mathrm{t}(3-\mathrm{t})\) \(0=6 \mathrm{t}(3-\mathrm{t}) \quad\{\because \mathrm{v}=0\}\) \(\mathrm{t} =0 \text { or } 3 \mathrm{sec}\) So, the particle will attain zero velocity at the time of 3 seconds. So, option (d) is correct.
AP EAMCET-20.08.2021
Motion in One Dimensions
141420
The slope of velocity - time graph for motion with uniform velocity is equal to:
1 Initial velocity
2 Final velocity
3 Zero
4 Constant velocity
Explanation:
C When velocity is uniform, velocity time graph is straight line parallel to the time axis. So, the slope is zero. For constant velocity acceleration, \(\mathrm{a}=0\) \(\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\) \(\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\) So, option (c) is correct
AP EAMCET-24.08.2021
Motion in One Dimensions
141421
The displacement of a particle moving along \(x\) axis is given by \(x=8 t^{2}+18 t\). The average acceleration during the interval \(t_{1}=2\) and \(t_{2}=\) \(4 \mathrm{~s}\) is
D According to equation, The displacement of the particle is given by, \(x =8 t^{2}+18 t\) \(\therefore v =\frac{d x}{d t}=16 t+18\) \(v =16 t+18\) at \(\mathrm{t}_{1}=2 \mathrm{sec} \Rightarrow \mathrm{v}_{1}=16 \times 2+18=32+18=50 \mathrm{~m} / \mathrm{s}\) at \(\mathrm{t}_{2}=4 \mathrm{sec} \Rightarrow \mathrm{v}_{2}=16 \times 4+18=64+18=82 \mathrm{~m} / \mathrm{s}\) average acceleration, \(\mathrm{a}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{t}}=\frac{82-50}{2}=\frac{32}{2}=16\) \(\mathrm{a}=16 \mathrm{~ms}^{-2}\) So, option (d) is correct
AP EAMCET-03.09.2021
Motion in One Dimensions
141422
The velocity-time graph of a moving object is shown in the figure. Find its maximum acceleration?
1 \(3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(6 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(4 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
B Acceleration in the \(1^{\text {st }}\) part. \(\mathrm{a}_{1}=\frac{20-0}{30-0}=0.66 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(2^{\text {nd }}\) part. \(\mathrm{a}_{2}=\frac{80-20}{40-30}=6 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(3^{\text {rd }}\) part. \(\mathrm{a}_{3}=\frac{0-80}{80-40}=-2 \mathrm{~m} / \mathrm{s}^{2}\) Max. acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Motion in One Dimensions
141418
The stopping distance of a moving vehicle is proportional to the
1 initial velocity
2 cube of the initial velocity
3 square of the initial velocity
4 cube root of the initial velocity
5 square root of the initial velocity
Explanation:
C The stopping distance of a moving vehicle is propotional to the square of the initial velocity. Let, Initial velocity \(=\mathrm{u}\) Final velocity \((v)=0\) Using third equation of motion \(v^{2}=u^{2}+2 a s\) \((0)^{2}=u^{2}+2 a s\) \(-u^{2}=-2 a s \quad \quad[\because a=\text { acceleration }\) \(\left.s=\frac{u^{2}}{2 a} \quad s=\text { stopping, side distance }\right]\) \(s \propto u^{2}\)
Kerala CEE 2021
Motion in One Dimensions
141419
The displacement of a particle starting from rest a \(t=0\) is given by \(s=9 t^2-2 t^3\). The time in seconds at which the particle will attain zero velocity is
1 \(8 \mathrm{~s}\)
2 \(6 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(3 \mathrm{~s}\)
Explanation:
D Given data, \(s=9 t^2-2 t^3\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=18 \mathrm{t}-6 \mathrm{t}^2\) \(\mathrm{v} =18 \mathrm{t}-6 \mathrm{t}^2\) \(0 =6 \mathrm{t}(3-\mathrm{t})\) \(0=6 \mathrm{t}(3-\mathrm{t}) \quad\{\because \mathrm{v}=0\}\) \(\mathrm{t} =0 \text { or } 3 \mathrm{sec}\) So, the particle will attain zero velocity at the time of 3 seconds. So, option (d) is correct.
AP EAMCET-20.08.2021
Motion in One Dimensions
141420
The slope of velocity - time graph for motion with uniform velocity is equal to:
1 Initial velocity
2 Final velocity
3 Zero
4 Constant velocity
Explanation:
C When velocity is uniform, velocity time graph is straight line parallel to the time axis. So, the slope is zero. For constant velocity acceleration, \(\mathrm{a}=0\) \(\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\) \(\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\) So, option (c) is correct
AP EAMCET-24.08.2021
Motion in One Dimensions
141421
The displacement of a particle moving along \(x\) axis is given by \(x=8 t^{2}+18 t\). The average acceleration during the interval \(t_{1}=2\) and \(t_{2}=\) \(4 \mathrm{~s}\) is
D According to equation, The displacement of the particle is given by, \(x =8 t^{2}+18 t\) \(\therefore v =\frac{d x}{d t}=16 t+18\) \(v =16 t+18\) at \(\mathrm{t}_{1}=2 \mathrm{sec} \Rightarrow \mathrm{v}_{1}=16 \times 2+18=32+18=50 \mathrm{~m} / \mathrm{s}\) at \(\mathrm{t}_{2}=4 \mathrm{sec} \Rightarrow \mathrm{v}_{2}=16 \times 4+18=64+18=82 \mathrm{~m} / \mathrm{s}\) average acceleration, \(\mathrm{a}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{t}}=\frac{82-50}{2}=\frac{32}{2}=16\) \(\mathrm{a}=16 \mathrm{~ms}^{-2}\) So, option (d) is correct
AP EAMCET-03.09.2021
Motion in One Dimensions
141422
The velocity-time graph of a moving object is shown in the figure. Find its maximum acceleration?
1 \(3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(6 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(4 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
B Acceleration in the \(1^{\text {st }}\) part. \(\mathrm{a}_{1}=\frac{20-0}{30-0}=0.66 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(2^{\text {nd }}\) part. \(\mathrm{a}_{2}=\frac{80-20}{40-30}=6 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(3^{\text {rd }}\) part. \(\mathrm{a}_{3}=\frac{0-80}{80-40}=-2 \mathrm{~m} / \mathrm{s}^{2}\) Max. acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\)
141418
The stopping distance of a moving vehicle is proportional to the
1 initial velocity
2 cube of the initial velocity
3 square of the initial velocity
4 cube root of the initial velocity
5 square root of the initial velocity
Explanation:
C The stopping distance of a moving vehicle is propotional to the square of the initial velocity. Let, Initial velocity \(=\mathrm{u}\) Final velocity \((v)=0\) Using third equation of motion \(v^{2}=u^{2}+2 a s\) \((0)^{2}=u^{2}+2 a s\) \(-u^{2}=-2 a s \quad \quad[\because a=\text { acceleration }\) \(\left.s=\frac{u^{2}}{2 a} \quad s=\text { stopping, side distance }\right]\) \(s \propto u^{2}\)
Kerala CEE 2021
Motion in One Dimensions
141419
The displacement of a particle starting from rest a \(t=0\) is given by \(s=9 t^2-2 t^3\). The time in seconds at which the particle will attain zero velocity is
1 \(8 \mathrm{~s}\)
2 \(6 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(3 \mathrm{~s}\)
Explanation:
D Given data, \(s=9 t^2-2 t^3\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=18 \mathrm{t}-6 \mathrm{t}^2\) \(\mathrm{v} =18 \mathrm{t}-6 \mathrm{t}^2\) \(0 =6 \mathrm{t}(3-\mathrm{t})\) \(0=6 \mathrm{t}(3-\mathrm{t}) \quad\{\because \mathrm{v}=0\}\) \(\mathrm{t} =0 \text { or } 3 \mathrm{sec}\) So, the particle will attain zero velocity at the time of 3 seconds. So, option (d) is correct.
AP EAMCET-20.08.2021
Motion in One Dimensions
141420
The slope of velocity - time graph for motion with uniform velocity is equal to:
1 Initial velocity
2 Final velocity
3 Zero
4 Constant velocity
Explanation:
C When velocity is uniform, velocity time graph is straight line parallel to the time axis. So, the slope is zero. For constant velocity acceleration, \(\mathrm{a}=0\) \(\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\) \(\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\) So, option (c) is correct
AP EAMCET-24.08.2021
Motion in One Dimensions
141421
The displacement of a particle moving along \(x\) axis is given by \(x=8 t^{2}+18 t\). The average acceleration during the interval \(t_{1}=2\) and \(t_{2}=\) \(4 \mathrm{~s}\) is
D According to equation, The displacement of the particle is given by, \(x =8 t^{2}+18 t\) \(\therefore v =\frac{d x}{d t}=16 t+18\) \(v =16 t+18\) at \(\mathrm{t}_{1}=2 \mathrm{sec} \Rightarrow \mathrm{v}_{1}=16 \times 2+18=32+18=50 \mathrm{~m} / \mathrm{s}\) at \(\mathrm{t}_{2}=4 \mathrm{sec} \Rightarrow \mathrm{v}_{2}=16 \times 4+18=64+18=82 \mathrm{~m} / \mathrm{s}\) average acceleration, \(\mathrm{a}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{t}}=\frac{82-50}{2}=\frac{32}{2}=16\) \(\mathrm{a}=16 \mathrm{~ms}^{-2}\) So, option (d) is correct
AP EAMCET-03.09.2021
Motion in One Dimensions
141422
The velocity-time graph of a moving object is shown in the figure. Find its maximum acceleration?
1 \(3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(6 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(4 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
B Acceleration in the \(1^{\text {st }}\) part. \(\mathrm{a}_{1}=\frac{20-0}{30-0}=0.66 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(2^{\text {nd }}\) part. \(\mathrm{a}_{2}=\frac{80-20}{40-30}=6 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(3^{\text {rd }}\) part. \(\mathrm{a}_{3}=\frac{0-80}{80-40}=-2 \mathrm{~m} / \mathrm{s}^{2}\) Max. acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\)
141418
The stopping distance of a moving vehicle is proportional to the
1 initial velocity
2 cube of the initial velocity
3 square of the initial velocity
4 cube root of the initial velocity
5 square root of the initial velocity
Explanation:
C The stopping distance of a moving vehicle is propotional to the square of the initial velocity. Let, Initial velocity \(=\mathrm{u}\) Final velocity \((v)=0\) Using third equation of motion \(v^{2}=u^{2}+2 a s\) \((0)^{2}=u^{2}+2 a s\) \(-u^{2}=-2 a s \quad \quad[\because a=\text { acceleration }\) \(\left.s=\frac{u^{2}}{2 a} \quad s=\text { stopping, side distance }\right]\) \(s \propto u^{2}\)
Kerala CEE 2021
Motion in One Dimensions
141419
The displacement of a particle starting from rest a \(t=0\) is given by \(s=9 t^2-2 t^3\). The time in seconds at which the particle will attain zero velocity is
1 \(8 \mathrm{~s}\)
2 \(6 \mathrm{~s}\)
3 \(4 \mathrm{~s}\)
4 \(3 \mathrm{~s}\)
Explanation:
D Given data, \(s=9 t^2-2 t^3\) \(\therefore \quad \mathrm{v} =\frac{\mathrm{ds}}{\mathrm{dt}}=18 \mathrm{t}-6 \mathrm{t}^2\) \(\mathrm{v} =18 \mathrm{t}-6 \mathrm{t}^2\) \(0 =6 \mathrm{t}(3-\mathrm{t})\) \(0=6 \mathrm{t}(3-\mathrm{t}) \quad\{\because \mathrm{v}=0\}\) \(\mathrm{t} =0 \text { or } 3 \mathrm{sec}\) So, the particle will attain zero velocity at the time of 3 seconds. So, option (d) is correct.
AP EAMCET-20.08.2021
Motion in One Dimensions
141420
The slope of velocity - time graph for motion with uniform velocity is equal to:
1 Initial velocity
2 Final velocity
3 Zero
4 Constant velocity
Explanation:
C When velocity is uniform, velocity time graph is straight line parallel to the time axis. So, the slope is zero. For constant velocity acceleration, \(\mathrm{a}=0\) \(\because \mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}\) \(\because \frac{\mathrm{dv}}{\mathrm{dt}}=0\) So, option (c) is correct
AP EAMCET-24.08.2021
Motion in One Dimensions
141421
The displacement of a particle moving along \(x\) axis is given by \(x=8 t^{2}+18 t\). The average acceleration during the interval \(t_{1}=2\) and \(t_{2}=\) \(4 \mathrm{~s}\) is
D According to equation, The displacement of the particle is given by, \(x =8 t^{2}+18 t\) \(\therefore v =\frac{d x}{d t}=16 t+18\) \(v =16 t+18\) at \(\mathrm{t}_{1}=2 \mathrm{sec} \Rightarrow \mathrm{v}_{1}=16 \times 2+18=32+18=50 \mathrm{~m} / \mathrm{s}\) at \(\mathrm{t}_{2}=4 \mathrm{sec} \Rightarrow \mathrm{v}_{2}=16 \times 4+18=64+18=82 \mathrm{~m} / \mathrm{s}\) average acceleration, \(\mathrm{a}=\frac{\mathrm{v}_{2}-\mathrm{v}_{1}}{\mathrm{t}}=\frac{82-50}{2}=\frac{32}{2}=16\) \(\mathrm{a}=16 \mathrm{~ms}^{-2}\) So, option (d) is correct
AP EAMCET-03.09.2021
Motion in One Dimensions
141422
The velocity-time graph of a moving object is shown in the figure. Find its maximum acceleration?
1 \(3 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
2 \(6 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(4 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
Explanation:
B Acceleration in the \(1^{\text {st }}\) part. \(\mathrm{a}_{1}=\frac{20-0}{30-0}=0.66 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(2^{\text {nd }}\) part. \(\mathrm{a}_{2}=\frac{80-20}{40-30}=6 \mathrm{~m} / \mathrm{s}^{2}\) Acceleration in the \(3^{\text {rd }}\) part. \(\mathrm{a}_{3}=\frac{0-80}{80-40}=-2 \mathrm{~m} / \mathrm{s}^{2}\) Max. acceleration is \(6 \mathrm{~m} / \mathrm{s}^{2}\)
