141494
Displacement \(x\) versus \(t^{2}\) graph is shown for a particle. The acceleration of the particle is \(\mathrm{x}(\mathrm{m})\)
1 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
A By given graph \(\mathrm{x}=\operatorname{slope}\left(\mathrm{t}^{2}\right)\) \(x=\frac{2-0}{1-0} \times t^{2}\) \(\mathrm{x}=2 \mathrm{t}^{2}\) Differentiating \(x\) w.r.t ' \(t\) ' we get velocity \(\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}\) Again differential \(\frac{\mathrm{dx}}{\mathrm{dt}}\) w.r.t ' \(\mathrm{t}\) ' we get acceleration \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4\) Which is the value of acceleration \(\text { i.e. } a=4 \mathrm{~m} / \mathrm{s}^{2}\)
UPSEE - 2017
Motion in One Dimensions
141497
A packet of weight ' \(W\) ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is
141498
The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between \(t=4 \mathrm{~s}\) and \(t\) \(=6 \mathrm{~s}\) is
1 \(10.5 \mathrm{~ms}^{-1}\)
2 \(12.5 \mathrm{~ms}^{-1}\)
3 \(7.5 \mathrm{~ms}^{-1}\)
4 \(9.5 \mathrm{~ms}^{-1}\)
Explanation:
B Given, time interval, \(t=t_{2}-t_{1}=2 \mathrm{sec}\) \(\mathrm{t}_{1}=4 \mathrm{sec}, \mathrm{t}_{2}=6 \mathrm{sec}\) slop of v-t graph, \([\mathrm{v}=\text { velocity }, \mathrm{t}=\text { time }]\) We know that, Acceleration, \(\mathrm{a}=\tan \theta=\frac{15}{6}=\frac{5}{2} \mathrm{~ms}^{-2}\) initial \(\mathrm{u}=0\) velocity at \(t_{1}=4 \mathrm{sec}\) then, \(\quad v=u+a t\) \(\mathrm{v} =0+\frac{5}{2} \times 4\) \(\mathrm{v}_{1} =5 \times 2=10 \mathrm{~m} / \mathrm{sec} \text { (given) }\) So, average \(\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{10+15}{2}=\frac{25}{2}=12.5=12.5 \mathrm{~m} / \mathrm{sec}\)
AP EAMCET-25.04.2017
Motion in One Dimensions
141499
The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
1 \(\frac{20}{7} \mathrm{~ms}^{-1}\)
2 \(\frac{18}{7} \mathrm{~ms}^{-1}\)
3 \(\frac{36}{7} \mathrm{~ms}^{-1}\)
4 \(\frac{12}{7} \mathrm{~ms}^{-1}\)
Explanation:
B \(\mathrm{V}_{\text {Avg }}=\frac{\text { Total distance }}{\text { total time }}\) Total distance \(=\mathrm{S}\) \(\mathrm{S}_{1}=\frac{1}{2} \times 2 \times 6=6 \mathrm{~m}\) \(\mathrm{~S}_{2}=6 \times 2=12\) \(\mathrm{~S}=\mathrm{S}_{1}+\mathrm{S}_{2}\) \(\quad=6+12=18 \mathrm{~m}\) \(\mathrm{V}_{\mathrm{Avg}}=\frac{18}{7} \mathrm{~m} / \mathrm{sec} .\)
141494
Displacement \(x\) versus \(t^{2}\) graph is shown for a particle. The acceleration of the particle is \(\mathrm{x}(\mathrm{m})\)
1 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
A By given graph \(\mathrm{x}=\operatorname{slope}\left(\mathrm{t}^{2}\right)\) \(x=\frac{2-0}{1-0} \times t^{2}\) \(\mathrm{x}=2 \mathrm{t}^{2}\) Differentiating \(x\) w.r.t ' \(t\) ' we get velocity \(\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}\) Again differential \(\frac{\mathrm{dx}}{\mathrm{dt}}\) w.r.t ' \(\mathrm{t}\) ' we get acceleration \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4\) Which is the value of acceleration \(\text { i.e. } a=4 \mathrm{~m} / \mathrm{s}^{2}\)
UPSEE - 2017
Motion in One Dimensions
141497
A packet of weight ' \(W\) ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is
141498
The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between \(t=4 \mathrm{~s}\) and \(t\) \(=6 \mathrm{~s}\) is
1 \(10.5 \mathrm{~ms}^{-1}\)
2 \(12.5 \mathrm{~ms}^{-1}\)
3 \(7.5 \mathrm{~ms}^{-1}\)
4 \(9.5 \mathrm{~ms}^{-1}\)
Explanation:
B Given, time interval, \(t=t_{2}-t_{1}=2 \mathrm{sec}\) \(\mathrm{t}_{1}=4 \mathrm{sec}, \mathrm{t}_{2}=6 \mathrm{sec}\) slop of v-t graph, \([\mathrm{v}=\text { velocity }, \mathrm{t}=\text { time }]\) We know that, Acceleration, \(\mathrm{a}=\tan \theta=\frac{15}{6}=\frac{5}{2} \mathrm{~ms}^{-2}\) initial \(\mathrm{u}=0\) velocity at \(t_{1}=4 \mathrm{sec}\) then, \(\quad v=u+a t\) \(\mathrm{v} =0+\frac{5}{2} \times 4\) \(\mathrm{v}_{1} =5 \times 2=10 \mathrm{~m} / \mathrm{sec} \text { (given) }\) So, average \(\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{10+15}{2}=\frac{25}{2}=12.5=12.5 \mathrm{~m} / \mathrm{sec}\)
AP EAMCET-25.04.2017
Motion in One Dimensions
141499
The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
1 \(\frac{20}{7} \mathrm{~ms}^{-1}\)
2 \(\frac{18}{7} \mathrm{~ms}^{-1}\)
3 \(\frac{36}{7} \mathrm{~ms}^{-1}\)
4 \(\frac{12}{7} \mathrm{~ms}^{-1}\)
Explanation:
B \(\mathrm{V}_{\text {Avg }}=\frac{\text { Total distance }}{\text { total time }}\) Total distance \(=\mathrm{S}\) \(\mathrm{S}_{1}=\frac{1}{2} \times 2 \times 6=6 \mathrm{~m}\) \(\mathrm{~S}_{2}=6 \times 2=12\) \(\mathrm{~S}=\mathrm{S}_{1}+\mathrm{S}_{2}\) \(\quad=6+12=18 \mathrm{~m}\) \(\mathrm{V}_{\mathrm{Avg}}=\frac{18}{7} \mathrm{~m} / \mathrm{sec} .\)
141494
Displacement \(x\) versus \(t^{2}\) graph is shown for a particle. The acceleration of the particle is \(\mathrm{x}(\mathrm{m})\)
1 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
A By given graph \(\mathrm{x}=\operatorname{slope}\left(\mathrm{t}^{2}\right)\) \(x=\frac{2-0}{1-0} \times t^{2}\) \(\mathrm{x}=2 \mathrm{t}^{2}\) Differentiating \(x\) w.r.t ' \(t\) ' we get velocity \(\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}\) Again differential \(\frac{\mathrm{dx}}{\mathrm{dt}}\) w.r.t ' \(\mathrm{t}\) ' we get acceleration \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4\) Which is the value of acceleration \(\text { i.e. } a=4 \mathrm{~m} / \mathrm{s}^{2}\)
UPSEE - 2017
Motion in One Dimensions
141497
A packet of weight ' \(W\) ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is
141498
The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between \(t=4 \mathrm{~s}\) and \(t\) \(=6 \mathrm{~s}\) is
1 \(10.5 \mathrm{~ms}^{-1}\)
2 \(12.5 \mathrm{~ms}^{-1}\)
3 \(7.5 \mathrm{~ms}^{-1}\)
4 \(9.5 \mathrm{~ms}^{-1}\)
Explanation:
B Given, time interval, \(t=t_{2}-t_{1}=2 \mathrm{sec}\) \(\mathrm{t}_{1}=4 \mathrm{sec}, \mathrm{t}_{2}=6 \mathrm{sec}\) slop of v-t graph, \([\mathrm{v}=\text { velocity }, \mathrm{t}=\text { time }]\) We know that, Acceleration, \(\mathrm{a}=\tan \theta=\frac{15}{6}=\frac{5}{2} \mathrm{~ms}^{-2}\) initial \(\mathrm{u}=0\) velocity at \(t_{1}=4 \mathrm{sec}\) then, \(\quad v=u+a t\) \(\mathrm{v} =0+\frac{5}{2} \times 4\) \(\mathrm{v}_{1} =5 \times 2=10 \mathrm{~m} / \mathrm{sec} \text { (given) }\) So, average \(\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{10+15}{2}=\frac{25}{2}=12.5=12.5 \mathrm{~m} / \mathrm{sec}\)
AP EAMCET-25.04.2017
Motion in One Dimensions
141499
The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
1 \(\frac{20}{7} \mathrm{~ms}^{-1}\)
2 \(\frac{18}{7} \mathrm{~ms}^{-1}\)
3 \(\frac{36}{7} \mathrm{~ms}^{-1}\)
4 \(\frac{12}{7} \mathrm{~ms}^{-1}\)
Explanation:
B \(\mathrm{V}_{\text {Avg }}=\frac{\text { Total distance }}{\text { total time }}\) Total distance \(=\mathrm{S}\) \(\mathrm{S}_{1}=\frac{1}{2} \times 2 \times 6=6 \mathrm{~m}\) \(\mathrm{~S}_{2}=6 \times 2=12\) \(\mathrm{~S}=\mathrm{S}_{1}+\mathrm{S}_{2}\) \(\quad=6+12=18 \mathrm{~m}\) \(\mathrm{V}_{\mathrm{Avg}}=\frac{18}{7} \mathrm{~m} / \mathrm{sec} .\)
141494
Displacement \(x\) versus \(t^{2}\) graph is shown for a particle. The acceleration of the particle is \(\mathrm{x}(\mathrm{m})\)
1 \(4 \mathrm{~m} / \mathrm{s}^{2}\)
2 \(8 \mathrm{~m} / \mathrm{s}^{2}\)
3 zero
4 \(2 \mathrm{~m} / \mathrm{s}^{2}\)
Explanation:
A By given graph \(\mathrm{x}=\operatorname{slope}\left(\mathrm{t}^{2}\right)\) \(x=\frac{2-0}{1-0} \times t^{2}\) \(\mathrm{x}=2 \mathrm{t}^{2}\) Differentiating \(x\) w.r.t ' \(t\) ' we get velocity \(\frac{\mathrm{dx}}{\mathrm{dt}}=4 \mathrm{t}\) Again differential \(\frac{\mathrm{dx}}{\mathrm{dt}}\) w.r.t ' \(\mathrm{t}\) ' we get acceleration \(\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=4\) Which is the value of acceleration \(\text { i.e. } a=4 \mathrm{~m} / \mathrm{s}^{2}\)
UPSEE - 2017
Motion in One Dimensions
141497
A packet of weight ' \(W\) ' dropped from a parachute strikes the ground and comes to rest with retardation equal to twice the acceleration due to gravity. The force exerted on the ground is
141498
The velocity and time graph for a particle moving in a straight line is shown in the figure. Then, the average velocity between \(t=4 \mathrm{~s}\) and \(t\) \(=6 \mathrm{~s}\) is
1 \(10.5 \mathrm{~ms}^{-1}\)
2 \(12.5 \mathrm{~ms}^{-1}\)
3 \(7.5 \mathrm{~ms}^{-1}\)
4 \(9.5 \mathrm{~ms}^{-1}\)
Explanation:
B Given, time interval, \(t=t_{2}-t_{1}=2 \mathrm{sec}\) \(\mathrm{t}_{1}=4 \mathrm{sec}, \mathrm{t}_{2}=6 \mathrm{sec}\) slop of v-t graph, \([\mathrm{v}=\text { velocity }, \mathrm{t}=\text { time }]\) We know that, Acceleration, \(\mathrm{a}=\tan \theta=\frac{15}{6}=\frac{5}{2} \mathrm{~ms}^{-2}\) initial \(\mathrm{u}=0\) velocity at \(t_{1}=4 \mathrm{sec}\) then, \(\quad v=u+a t\) \(\mathrm{v} =0+\frac{5}{2} \times 4\) \(\mathrm{v}_{1} =5 \times 2=10 \mathrm{~m} / \mathrm{sec} \text { (given) }\) So, average \(\frac{\mathrm{v}_{1}+\mathrm{v}_{2}}{2}=\frac{10+15}{2}=\frac{25}{2}=12.5=12.5 \mathrm{~m} / \mathrm{sec}\)
AP EAMCET-25.04.2017
Motion in One Dimensions
141499
The variation of velocity (v) of a particle with time (t) is shown in the figure. The average velocity of the particle during its motion is
1 \(\frac{20}{7} \mathrm{~ms}^{-1}\)
2 \(\frac{18}{7} \mathrm{~ms}^{-1}\)
3 \(\frac{36}{7} \mathrm{~ms}^{-1}\)
4 \(\frac{12}{7} \mathrm{~ms}^{-1}\)
Explanation:
B \(\mathrm{V}_{\text {Avg }}=\frac{\text { Total distance }}{\text { total time }}\) Total distance \(=\mathrm{S}\) \(\mathrm{S}_{1}=\frac{1}{2} \times 2 \times 6=6 \mathrm{~m}\) \(\mathrm{~S}_{2}=6 \times 2=12\) \(\mathrm{~S}=\mathrm{S}_{1}+\mathrm{S}_{2}\) \(\quad=6+12=18 \mathrm{~m}\) \(\mathrm{V}_{\mathrm{Avg}}=\frac{18}{7} \mathrm{~m} / \mathrm{sec} .\)
