141467
The given graph shows the variation of velocity \(v\) with position \(x\) for a particle moving along a straight line : Which of the following graph shown the variation of acceleration a with position \(x\) ?
1
2
3
4
Explanation:
C According to velocity-position graph, \(\mathrm{v}=-\mathrm{mx}+\mathrm{v}_{0}\) Where, \(\mathrm{m}=\) Slope of graph which is \(-\mathrm{ve}\) \(\mathrm{v}_{0}=\) Intercept of line which is \(+\mathrm{ve}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From equation (1) \(a=-m\left(-m x+v_{0}\right)\) \(a=m^{2} x-m v_{0} \ldots \ldots(2)\) Slope \(=+\mathrm{ve}\) Intercept \(=-\mathrm{ve}\)
Karnataka CET-2019
Motion in One Dimensions
141469
Consider the following velocity and time graph: Which one of the following is the value of average acceleration from \(8 \mathrm{~s}\) to \(12 \mathrm{~s}\) ?
1 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(12 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(-1 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
D given that, Average acceleration \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4-8}{12-8}=\frac{-4}{4}\) \(\mathrm{a}=-1 \mathrm{~m} / \mathrm{sec}^{2}\)
NDA (II) 2018
Motion in One Dimensions
141470
The relation between time \(\mathbf{t}\) and distance \(\mathbf{x}\) for a moving particle is \(t=\alpha x^{2}+\beta x\), where \(\alpha\) and \(\beta\) are constants. If \(v\) is the velocity at distance \(x\), then the retardation of the particle is
1 \(2 \alpha v^{3}\)
2 \(2 \beta \mathrm{v}^{3}\)
3 \(2 \alpha \beta v^{3}\)
4 \(2 \beta^{2} v^{2}\)
Explanation:
A Given, \(t=\alpha x^{2}+\beta x\) Differentiating w.r.t to time, we have \(1=2 \alpha x \frac{d x}{d t}+\beta \frac{d x}{d t}\) We know, \(1=2 \alpha \mathrm{xv}+\beta \mathrm{v}\) \(\mathrm{v}(2 \alpha \mathrm{x}+\beta)=1\) Again differentiating equation (i) w.r.t to time, we have \(0=2 \alpha x \frac{d^{2} x}{d t^{2}}+\frac{\beta d^{2} x}{d t^{2}}+2 \alpha\left(\frac{d x}{d t}\right)^{2}\) \(0=(2 \alpha x+\beta) a+2 \alpha v^{2} \quad\left(\because \frac{d^{2} x}{d t^{2}}=a\right)\) \(a=-2 \alpha v^{3} m / \sec ^{2}\) \(\therefore\) Retardation \(=2 \alpha \mathrm{v}^{3} \mathrm{~m} / \mathrm{sec}^{2}\)
AMU-2018
Motion in One Dimensions
141471
Acceleration-time graph of a body moving in a straight line is as shown. The body started its motion from rest. At which point is the body moving with the largest speed?
1 1
2 2
3 3
4 4
Explanation:
B According the acceleration - time graph, speed \(=\) area under acceleration - time graph . As given in the graph area is maximum for point 0 to 2 . Therefore, At point 2, the body moves with the largest speed.
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Motion in One Dimensions
141467
The given graph shows the variation of velocity \(v\) with position \(x\) for a particle moving along a straight line : Which of the following graph shown the variation of acceleration a with position \(x\) ?
1
2
3
4
Explanation:
C According to velocity-position graph, \(\mathrm{v}=-\mathrm{mx}+\mathrm{v}_{0}\) Where, \(\mathrm{m}=\) Slope of graph which is \(-\mathrm{ve}\) \(\mathrm{v}_{0}=\) Intercept of line which is \(+\mathrm{ve}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From equation (1) \(a=-m\left(-m x+v_{0}\right)\) \(a=m^{2} x-m v_{0} \ldots \ldots(2)\) Slope \(=+\mathrm{ve}\) Intercept \(=-\mathrm{ve}\)
Karnataka CET-2019
Motion in One Dimensions
141469
Consider the following velocity and time graph: Which one of the following is the value of average acceleration from \(8 \mathrm{~s}\) to \(12 \mathrm{~s}\) ?
1 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(12 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(-1 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
D given that, Average acceleration \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4-8}{12-8}=\frac{-4}{4}\) \(\mathrm{a}=-1 \mathrm{~m} / \mathrm{sec}^{2}\)
NDA (II) 2018
Motion in One Dimensions
141470
The relation between time \(\mathbf{t}\) and distance \(\mathbf{x}\) for a moving particle is \(t=\alpha x^{2}+\beta x\), where \(\alpha\) and \(\beta\) are constants. If \(v\) is the velocity at distance \(x\), then the retardation of the particle is
1 \(2 \alpha v^{3}\)
2 \(2 \beta \mathrm{v}^{3}\)
3 \(2 \alpha \beta v^{3}\)
4 \(2 \beta^{2} v^{2}\)
Explanation:
A Given, \(t=\alpha x^{2}+\beta x\) Differentiating w.r.t to time, we have \(1=2 \alpha x \frac{d x}{d t}+\beta \frac{d x}{d t}\) We know, \(1=2 \alpha \mathrm{xv}+\beta \mathrm{v}\) \(\mathrm{v}(2 \alpha \mathrm{x}+\beta)=1\) Again differentiating equation (i) w.r.t to time, we have \(0=2 \alpha x \frac{d^{2} x}{d t^{2}}+\frac{\beta d^{2} x}{d t^{2}}+2 \alpha\left(\frac{d x}{d t}\right)^{2}\) \(0=(2 \alpha x+\beta) a+2 \alpha v^{2} \quad\left(\because \frac{d^{2} x}{d t^{2}}=a\right)\) \(a=-2 \alpha v^{3} m / \sec ^{2}\) \(\therefore\) Retardation \(=2 \alpha \mathrm{v}^{3} \mathrm{~m} / \mathrm{sec}^{2}\)
AMU-2018
Motion in One Dimensions
141471
Acceleration-time graph of a body moving in a straight line is as shown. The body started its motion from rest. At which point is the body moving with the largest speed?
1 1
2 2
3 3
4 4
Explanation:
B According the acceleration - time graph, speed \(=\) area under acceleration - time graph . As given in the graph area is maximum for point 0 to 2 . Therefore, At point 2, the body moves with the largest speed.
141467
The given graph shows the variation of velocity \(v\) with position \(x\) for a particle moving along a straight line : Which of the following graph shown the variation of acceleration a with position \(x\) ?
1
2
3
4
Explanation:
C According to velocity-position graph, \(\mathrm{v}=-\mathrm{mx}+\mathrm{v}_{0}\) Where, \(\mathrm{m}=\) Slope of graph which is \(-\mathrm{ve}\) \(\mathrm{v}_{0}=\) Intercept of line which is \(+\mathrm{ve}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From equation (1) \(a=-m\left(-m x+v_{0}\right)\) \(a=m^{2} x-m v_{0} \ldots \ldots(2)\) Slope \(=+\mathrm{ve}\) Intercept \(=-\mathrm{ve}\)
Karnataka CET-2019
Motion in One Dimensions
141469
Consider the following velocity and time graph: Which one of the following is the value of average acceleration from \(8 \mathrm{~s}\) to \(12 \mathrm{~s}\) ?
1 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(12 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(-1 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
D given that, Average acceleration \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4-8}{12-8}=\frac{-4}{4}\) \(\mathrm{a}=-1 \mathrm{~m} / \mathrm{sec}^{2}\)
NDA (II) 2018
Motion in One Dimensions
141470
The relation between time \(\mathbf{t}\) and distance \(\mathbf{x}\) for a moving particle is \(t=\alpha x^{2}+\beta x\), where \(\alpha\) and \(\beta\) are constants. If \(v\) is the velocity at distance \(x\), then the retardation of the particle is
1 \(2 \alpha v^{3}\)
2 \(2 \beta \mathrm{v}^{3}\)
3 \(2 \alpha \beta v^{3}\)
4 \(2 \beta^{2} v^{2}\)
Explanation:
A Given, \(t=\alpha x^{2}+\beta x\) Differentiating w.r.t to time, we have \(1=2 \alpha x \frac{d x}{d t}+\beta \frac{d x}{d t}\) We know, \(1=2 \alpha \mathrm{xv}+\beta \mathrm{v}\) \(\mathrm{v}(2 \alpha \mathrm{x}+\beta)=1\) Again differentiating equation (i) w.r.t to time, we have \(0=2 \alpha x \frac{d^{2} x}{d t^{2}}+\frac{\beta d^{2} x}{d t^{2}}+2 \alpha\left(\frac{d x}{d t}\right)^{2}\) \(0=(2 \alpha x+\beta) a+2 \alpha v^{2} \quad\left(\because \frac{d^{2} x}{d t^{2}}=a\right)\) \(a=-2 \alpha v^{3} m / \sec ^{2}\) \(\therefore\) Retardation \(=2 \alpha \mathrm{v}^{3} \mathrm{~m} / \mathrm{sec}^{2}\)
AMU-2018
Motion in One Dimensions
141471
Acceleration-time graph of a body moving in a straight line is as shown. The body started its motion from rest. At which point is the body moving with the largest speed?
1 1
2 2
3 3
4 4
Explanation:
B According the acceleration - time graph, speed \(=\) area under acceleration - time graph . As given in the graph area is maximum for point 0 to 2 . Therefore, At point 2, the body moves with the largest speed.
141467
The given graph shows the variation of velocity \(v\) with position \(x\) for a particle moving along a straight line : Which of the following graph shown the variation of acceleration a with position \(x\) ?
1
2
3
4
Explanation:
C According to velocity-position graph, \(\mathrm{v}=-\mathrm{mx}+\mathrm{v}_{0}\) Where, \(\mathrm{m}=\) Slope of graph which is \(-\mathrm{ve}\) \(\mathrm{v}_{0}=\) Intercept of line which is \(+\mathrm{ve}\) \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From equation (1) \(a=-m\left(-m x+v_{0}\right)\) \(a=m^{2} x-m v_{0} \ldots \ldots(2)\) Slope \(=+\mathrm{ve}\) Intercept \(=-\mathrm{ve}\)
Karnataka CET-2019
Motion in One Dimensions
141469
Consider the following velocity and time graph: Which one of the following is the value of average acceleration from \(8 \mathrm{~s}\) to \(12 \mathrm{~s}\) ?
1 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(12 \mathrm{~m} / \mathrm{s}^{2}\)
3 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
4 \(-1 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
D given that, Average acceleration \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{4-8}{12-8}=\frac{-4}{4}\) \(\mathrm{a}=-1 \mathrm{~m} / \mathrm{sec}^{2}\)
NDA (II) 2018
Motion in One Dimensions
141470
The relation between time \(\mathbf{t}\) and distance \(\mathbf{x}\) for a moving particle is \(t=\alpha x^{2}+\beta x\), where \(\alpha\) and \(\beta\) are constants. If \(v\) is the velocity at distance \(x\), then the retardation of the particle is
1 \(2 \alpha v^{3}\)
2 \(2 \beta \mathrm{v}^{3}\)
3 \(2 \alpha \beta v^{3}\)
4 \(2 \beta^{2} v^{2}\)
Explanation:
A Given, \(t=\alpha x^{2}+\beta x\) Differentiating w.r.t to time, we have \(1=2 \alpha x \frac{d x}{d t}+\beta \frac{d x}{d t}\) We know, \(1=2 \alpha \mathrm{xv}+\beta \mathrm{v}\) \(\mathrm{v}(2 \alpha \mathrm{x}+\beta)=1\) Again differentiating equation (i) w.r.t to time, we have \(0=2 \alpha x \frac{d^{2} x}{d t^{2}}+\frac{\beta d^{2} x}{d t^{2}}+2 \alpha\left(\frac{d x}{d t}\right)^{2}\) \(0=(2 \alpha x+\beta) a+2 \alpha v^{2} \quad\left(\because \frac{d^{2} x}{d t^{2}}=a\right)\) \(a=-2 \alpha v^{3} m / \sec ^{2}\) \(\therefore\) Retardation \(=2 \alpha \mathrm{v}^{3} \mathrm{~m} / \mathrm{sec}^{2}\)
AMU-2018
Motion in One Dimensions
141471
Acceleration-time graph of a body moving in a straight line is as shown. The body started its motion from rest. At which point is the body moving with the largest speed?
1 1
2 2
3 3
4 4
Explanation:
B According the acceleration - time graph, speed \(=\) area under acceleration - time graph . As given in the graph area is maximum for point 0 to 2 . Therefore, At point 2, the body moves with the largest speed.
