141380
In 1s, a particle goes from point \(A\) to point \(B\) moving in a semi-circle of radius \(1 \mathrm{~m}\) as shown in the figure. The magnitude of average velocity is
1 \(3.14 \mathrm{~m} / \mathrm{s}\)
2 \(2 \mathrm{~m} / \mathrm{s}\)
3 \(1 \mathrm{~m} / \mathrm{s}\)
4 0
Explanation:
B Total displacement \(=\) shortest distance \(=\) diameter of semicircle \(=2 \mathrm{~m}\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\) \(=\frac{2}{1}=2 \mathrm{~m} / \mathrm{s}\)
Assam CEE-2014
Motion in One Dimensions
141381
The figure below shows the variation of velocity \(v\) of a body with position \(x\) from the origin \(O\). Which of the answer of the answer figures correctly represents the variation of acceleration \(\alpha\) with position \(x\) ?
1
2
3
4
Explanation:
D \(\mathrm{V}=\mathrm{mx}+\mathrm{v}_{0} \quad \ldots\) (i) \(\mathrm{m}\) - slope of graph which is negative \(v_{0}\) - intercept of line which is positive \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From (i) \(\mathrm{a}=-\mathrm{m}\left(-\mathrm{mx}+\mathrm{v}_{0}\right)\) \(\mathrm{a}=\mathrm{m}^{2} \mathrm{x}-\mathrm{mv}_{0}\) Slope of (+ve) and Intercept is negative
Assam CEE-2014
Motion in One Dimensions
141382
A car starting from rest is accelerated at a constant rate \(\alpha\) until it attains a speed \(v\). It is then retarded at a constant rate \(\beta\) until it comes to rest. The average speed of the car during its entire journey is
1 0
2 \(\frac{v}{2}\)
3 \(\frac{\alpha v}{2 \beta}\)
4 \(\frac{\beta v}{2 \alpha}\)
Explanation:
B Given that Acceleration of the car \(=\propto\) The average speed of the car \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}}{\mathrm{t}}\) When it start at rest \(u=0\) in time \(t_{1}\) \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(\mathrm{v}=\mathrm{at}_{1}\) And distance travelled \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}\) \(\mathrm{~s}_{1}=\frac{1}{2} \mathrm{at}_{1}^{2}\) When it comes to rate in time \(t_{2}\) Rate of acceleration \(=\) Rate of deacceleration When \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{1}{2} \mathrm{at}_{2}^{2}\) Since \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{\mathrm{at}_{1}^{2}}{2}\) From eq \(^{\text {n }}\) (i) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}_{1}+\mathrm{s}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\frac{\frac{\mathrm{at}_{1}^{2}}{2}+\frac{\mathrm{at}_{1}^{2}}{2}}{2 \mathrm{t}_{1}}\) From eq \(^{\mathrm{n}}\) (i) \(\frac{2 \mathrm{at}_{1}^{2}}{2 \times 2 \mathrm{t}_{1}}=\frac{\mathrm{at}_{1}}{2}\) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{v}}{2}\)
Assam CEE-2014
Motion in One Dimensions
141383
Figure given shows the distance-time graph of the motion of a car. It follows from the graph that the car is
1 at rest
2 in uniform motion
3 in non-uniform acceleration
4 uniformly accelerated
Explanation:
D The curve of \(x, t\) of car motion is parabolic- \(\mathrm{x}=1.2 \mathrm{t}^{2}\) Then, velocity \((v)=\frac{d x}{d t}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(1.2 \mathrm{t}^{2}\right)\) \(\mathrm{v} =2.4 \mathrm{t}\) And, Acceleration \((\mathrm{a})=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(2.4 \mathrm{t})\) \(a=2.4 \quad(\text { constant })\) So, the motion of car is uniform acceleration.
141380
In 1s, a particle goes from point \(A\) to point \(B\) moving in a semi-circle of radius \(1 \mathrm{~m}\) as shown in the figure. The magnitude of average velocity is
1 \(3.14 \mathrm{~m} / \mathrm{s}\)
2 \(2 \mathrm{~m} / \mathrm{s}\)
3 \(1 \mathrm{~m} / \mathrm{s}\)
4 0
Explanation:
B Total displacement \(=\) shortest distance \(=\) diameter of semicircle \(=2 \mathrm{~m}\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\) \(=\frac{2}{1}=2 \mathrm{~m} / \mathrm{s}\)
Assam CEE-2014
Motion in One Dimensions
141381
The figure below shows the variation of velocity \(v\) of a body with position \(x\) from the origin \(O\). Which of the answer of the answer figures correctly represents the variation of acceleration \(\alpha\) with position \(x\) ?
1
2
3
4
Explanation:
D \(\mathrm{V}=\mathrm{mx}+\mathrm{v}_{0} \quad \ldots\) (i) \(\mathrm{m}\) - slope of graph which is negative \(v_{0}\) - intercept of line which is positive \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From (i) \(\mathrm{a}=-\mathrm{m}\left(-\mathrm{mx}+\mathrm{v}_{0}\right)\) \(\mathrm{a}=\mathrm{m}^{2} \mathrm{x}-\mathrm{mv}_{0}\) Slope of (+ve) and Intercept is negative
Assam CEE-2014
Motion in One Dimensions
141382
A car starting from rest is accelerated at a constant rate \(\alpha\) until it attains a speed \(v\). It is then retarded at a constant rate \(\beta\) until it comes to rest. The average speed of the car during its entire journey is
1 0
2 \(\frac{v}{2}\)
3 \(\frac{\alpha v}{2 \beta}\)
4 \(\frac{\beta v}{2 \alpha}\)
Explanation:
B Given that Acceleration of the car \(=\propto\) The average speed of the car \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}}{\mathrm{t}}\) When it start at rest \(u=0\) in time \(t_{1}\) \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(\mathrm{v}=\mathrm{at}_{1}\) And distance travelled \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}\) \(\mathrm{~s}_{1}=\frac{1}{2} \mathrm{at}_{1}^{2}\) When it comes to rate in time \(t_{2}\) Rate of acceleration \(=\) Rate of deacceleration When \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{1}{2} \mathrm{at}_{2}^{2}\) Since \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{\mathrm{at}_{1}^{2}}{2}\) From eq \(^{\text {n }}\) (i) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}_{1}+\mathrm{s}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\frac{\frac{\mathrm{at}_{1}^{2}}{2}+\frac{\mathrm{at}_{1}^{2}}{2}}{2 \mathrm{t}_{1}}\) From eq \(^{\mathrm{n}}\) (i) \(\frac{2 \mathrm{at}_{1}^{2}}{2 \times 2 \mathrm{t}_{1}}=\frac{\mathrm{at}_{1}}{2}\) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{v}}{2}\)
Assam CEE-2014
Motion in One Dimensions
141383
Figure given shows the distance-time graph of the motion of a car. It follows from the graph that the car is
1 at rest
2 in uniform motion
3 in non-uniform acceleration
4 uniformly accelerated
Explanation:
D The curve of \(x, t\) of car motion is parabolic- \(\mathrm{x}=1.2 \mathrm{t}^{2}\) Then, velocity \((v)=\frac{d x}{d t}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(1.2 \mathrm{t}^{2}\right)\) \(\mathrm{v} =2.4 \mathrm{t}\) And, Acceleration \((\mathrm{a})=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(2.4 \mathrm{t})\) \(a=2.4 \quad(\text { constant })\) So, the motion of car is uniform acceleration.
141380
In 1s, a particle goes from point \(A\) to point \(B\) moving in a semi-circle of radius \(1 \mathrm{~m}\) as shown in the figure. The magnitude of average velocity is
1 \(3.14 \mathrm{~m} / \mathrm{s}\)
2 \(2 \mathrm{~m} / \mathrm{s}\)
3 \(1 \mathrm{~m} / \mathrm{s}\)
4 0
Explanation:
B Total displacement \(=\) shortest distance \(=\) diameter of semicircle \(=2 \mathrm{~m}\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\) \(=\frac{2}{1}=2 \mathrm{~m} / \mathrm{s}\)
Assam CEE-2014
Motion in One Dimensions
141381
The figure below shows the variation of velocity \(v\) of a body with position \(x\) from the origin \(O\). Which of the answer of the answer figures correctly represents the variation of acceleration \(\alpha\) with position \(x\) ?
1
2
3
4
Explanation:
D \(\mathrm{V}=\mathrm{mx}+\mathrm{v}_{0} \quad \ldots\) (i) \(\mathrm{m}\) - slope of graph which is negative \(v_{0}\) - intercept of line which is positive \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From (i) \(\mathrm{a}=-\mathrm{m}\left(-\mathrm{mx}+\mathrm{v}_{0}\right)\) \(\mathrm{a}=\mathrm{m}^{2} \mathrm{x}-\mathrm{mv}_{0}\) Slope of (+ve) and Intercept is negative
Assam CEE-2014
Motion in One Dimensions
141382
A car starting from rest is accelerated at a constant rate \(\alpha\) until it attains a speed \(v\). It is then retarded at a constant rate \(\beta\) until it comes to rest. The average speed of the car during its entire journey is
1 0
2 \(\frac{v}{2}\)
3 \(\frac{\alpha v}{2 \beta}\)
4 \(\frac{\beta v}{2 \alpha}\)
Explanation:
B Given that Acceleration of the car \(=\propto\) The average speed of the car \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}}{\mathrm{t}}\) When it start at rest \(u=0\) in time \(t_{1}\) \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(\mathrm{v}=\mathrm{at}_{1}\) And distance travelled \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}\) \(\mathrm{~s}_{1}=\frac{1}{2} \mathrm{at}_{1}^{2}\) When it comes to rate in time \(t_{2}\) Rate of acceleration \(=\) Rate of deacceleration When \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{1}{2} \mathrm{at}_{2}^{2}\) Since \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{\mathrm{at}_{1}^{2}}{2}\) From eq \(^{\text {n }}\) (i) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}_{1}+\mathrm{s}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\frac{\frac{\mathrm{at}_{1}^{2}}{2}+\frac{\mathrm{at}_{1}^{2}}{2}}{2 \mathrm{t}_{1}}\) From eq \(^{\mathrm{n}}\) (i) \(\frac{2 \mathrm{at}_{1}^{2}}{2 \times 2 \mathrm{t}_{1}}=\frac{\mathrm{at}_{1}}{2}\) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{v}}{2}\)
Assam CEE-2014
Motion in One Dimensions
141383
Figure given shows the distance-time graph of the motion of a car. It follows from the graph that the car is
1 at rest
2 in uniform motion
3 in non-uniform acceleration
4 uniformly accelerated
Explanation:
D The curve of \(x, t\) of car motion is parabolic- \(\mathrm{x}=1.2 \mathrm{t}^{2}\) Then, velocity \((v)=\frac{d x}{d t}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(1.2 \mathrm{t}^{2}\right)\) \(\mathrm{v} =2.4 \mathrm{t}\) And, Acceleration \((\mathrm{a})=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(2.4 \mathrm{t})\) \(a=2.4 \quad(\text { constant })\) So, the motion of car is uniform acceleration.
141380
In 1s, a particle goes from point \(A\) to point \(B\) moving in a semi-circle of radius \(1 \mathrm{~m}\) as shown in the figure. The magnitude of average velocity is
1 \(3.14 \mathrm{~m} / \mathrm{s}\)
2 \(2 \mathrm{~m} / \mathrm{s}\)
3 \(1 \mathrm{~m} / \mathrm{s}\)
4 0
Explanation:
B Total displacement \(=\) shortest distance \(=\) diameter of semicircle \(=2 \mathrm{~m}\) Average velocity \(=\frac{\text { Total displacement }}{\text { Total time }}\) \(=\frac{2}{1}=2 \mathrm{~m} / \mathrm{s}\)
Assam CEE-2014
Motion in One Dimensions
141381
The figure below shows the variation of velocity \(v\) of a body with position \(x\) from the origin \(O\). Which of the answer of the answer figures correctly represents the variation of acceleration \(\alpha\) with position \(x\) ?
1
2
3
4
Explanation:
D \(\mathrm{V}=\mathrm{mx}+\mathrm{v}_{0} \quad \ldots\) (i) \(\mathrm{m}\) - slope of graph which is negative \(v_{0}\) - intercept of line which is positive \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}}=-\mathrm{m}\left(\frac{\mathrm{dx}}{\mathrm{dt}}\right)+\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{v}_{0}\right)\) \(a=-m v+0\) From (i) \(\mathrm{a}=-\mathrm{m}\left(-\mathrm{mx}+\mathrm{v}_{0}\right)\) \(\mathrm{a}=\mathrm{m}^{2} \mathrm{x}-\mathrm{mv}_{0}\) Slope of (+ve) and Intercept is negative
Assam CEE-2014
Motion in One Dimensions
141382
A car starting from rest is accelerated at a constant rate \(\alpha\) until it attains a speed \(v\). It is then retarded at a constant rate \(\beta\) until it comes to rest. The average speed of the car during its entire journey is
1 0
2 \(\frac{v}{2}\)
3 \(\frac{\alpha v}{2 \beta}\)
4 \(\frac{\beta v}{2 \alpha}\)
Explanation:
B Given that Acceleration of the car \(=\propto\) The average speed of the car \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}}{\mathrm{t}}\) When it start at rest \(u=0\) in time \(t_{1}\) \(\mathrm{v}=\mathrm{u}+\mathrm{at}\) \(\mathrm{v}=\mathrm{at}_{1}\) And distance travelled \(\mathrm{s}=\mathrm{ut}+\frac{1}{2} a \mathrm{t}^{2}\) \(\mathrm{~s}_{1}=\frac{1}{2} \mathrm{at}_{1}^{2}\) When it comes to rate in time \(t_{2}\) Rate of acceleration \(=\) Rate of deacceleration When \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{1}{2} \mathrm{at}_{2}^{2}\) Since \(\mathrm{t}_{1}=\mathrm{t}_{2} \quad \mathrm{~s}_{2}=\frac{\mathrm{at}_{1}^{2}}{2}\) From eq \(^{\text {n }}\) (i) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{s}_{1}+\mathrm{s}_{2}}{\mathrm{t}_{1}+\mathrm{t}_{2}}=\frac{\frac{\mathrm{at}_{1}^{2}}{2}+\frac{\mathrm{at}_{1}^{2}}{2}}{2 \mathrm{t}_{1}}\) From eq \(^{\mathrm{n}}\) (i) \(\frac{2 \mathrm{at}_{1}^{2}}{2 \times 2 \mathrm{t}_{1}}=\frac{\mathrm{at}_{1}}{2}\) \(\mathrm{v}_{\mathrm{avg}}=\frac{\mathrm{v}}{2}\)
Assam CEE-2014
Motion in One Dimensions
141383
Figure given shows the distance-time graph of the motion of a car. It follows from the graph that the car is
1 at rest
2 in uniform motion
3 in non-uniform acceleration
4 uniformly accelerated
Explanation:
D The curve of \(x, t\) of car motion is parabolic- \(\mathrm{x}=1.2 \mathrm{t}^{2}\) Then, velocity \((v)=\frac{d x}{d t}\) \(\mathrm{v} =\frac{\mathrm{d}}{\mathrm{dt}}\left(1.2 \mathrm{t}^{2}\right)\) \(\mathrm{v} =2.4 \mathrm{t}\) And, Acceleration \((\mathrm{a})=\frac{\mathrm{dv}}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}(2.4 \mathrm{t})\) \(a=2.4 \quad(\text { constant })\) So, the motion of car is uniform acceleration.
