141389
If \(d\) denotes the distance covered by a car in time \(t\) and \(\vec{S}\) denotes the displacement by the car during the same time, then:
1 \(d \leq|\vec{S}|\)
2 \(d=|\vec{S}|\)
3 \(d \geq|\vec{S}|\)
4 \(d \lt |\vec{S}|\)
Explanation:
C Given, If \(d\) denotes the distance covered by a car in time \(t\) and \(\overrightarrow{\mathrm{S}}\) denotes the displacement by the car during the same that, Displacement is the shortest distance covered by a body and distance is actual path covered by a moving object. Therefore, \(\mathrm{A}-\mathrm{B}-\mathrm{C}\) distance A \(-\mathrm{C}\) displacement \(\mathrm{d} \geq|\overrightarrow{\mathrm{S}}|\)
NDA (I) 2013
Motion in One Dimensions
141390 The motion of a particle is given by a straight line in the graph given above drawn with displacement \((x)\) and time ( \(t\) ). Which one among the following statements is correct?
1 The velocity of the particle is uniform
2 The velocity of the particle is non-uniform
3 The speed is uniform and the particle is moving on a circular path
4 The speed is non-uniform and the particle is moving on a straight line path
Explanation:
A See above graph. The velocity of the particle is uniform
NDA (I) 2013
Motion in One Dimensions
141391 The plot given above represents displacement ' \(x\) ' of a particle with time ' \(t\) '. The particle is
1 moving with uniform velocity
2 moving with acceleration
3 moving with deceleration
4 executing a periodic motion
Explanation:
B The above given graph is an upward parabola. \(s=u t+\frac{1}{2} a t^{2}\) The above equation of motion is also a parabolic equation. For the upward parabolic equation the \(\mathrm{t}^{2}\) term should be positive. Hence acceleration of the particle is positive. Thus the particle is moving with acceleration. (b) Speed, Velocity and Acceleration
NDA (II) 2013
Motion in One Dimensions
141392
The position \(x\) of a particle varies with time \((t)\) as \(x=\mathbf{A t}^{2}-\mathbf{B} \mathbf{t}^{3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\) ?
1 \(\frac{2 \mathrm{~A}}{3 \mathrm{~B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{\mathrm{A}}{3 \mathrm{~B}}\)
4 zero
Explanation:
C Given that, \(\mathrm{x}=\mathrm{At}^{2}-\mathrm{Bt}^{3}\) \((v)=\frac{d x}{d t}=2 A t-3 B t^{2}\) Acceleration \((\mathrm{a})=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{At}-3 \mathrm{Bt}^{2}\right)\) \(\mathrm{a}=2 \mathrm{~A}-6 \mathrm{Bt}\) Put \(\mathrm{a}=0\) \(2 \mathrm{~A}=6 \mathrm{Bt}\) \(\mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}\) \(\mathrm{t}=\frac{\mathrm{A}}{3 \mathrm{~B}}\)
141389
If \(d\) denotes the distance covered by a car in time \(t\) and \(\vec{S}\) denotes the displacement by the car during the same time, then:
1 \(d \leq|\vec{S}|\)
2 \(d=|\vec{S}|\)
3 \(d \geq|\vec{S}|\)
4 \(d \lt |\vec{S}|\)
Explanation:
C Given, If \(d\) denotes the distance covered by a car in time \(t\) and \(\overrightarrow{\mathrm{S}}\) denotes the displacement by the car during the same that, Displacement is the shortest distance covered by a body and distance is actual path covered by a moving object. Therefore, \(\mathrm{A}-\mathrm{B}-\mathrm{C}\) distance A \(-\mathrm{C}\) displacement \(\mathrm{d} \geq|\overrightarrow{\mathrm{S}}|\)
NDA (I) 2013
Motion in One Dimensions
141390 The motion of a particle is given by a straight line in the graph given above drawn with displacement \((x)\) and time ( \(t\) ). Which one among the following statements is correct?
1 The velocity of the particle is uniform
2 The velocity of the particle is non-uniform
3 The speed is uniform and the particle is moving on a circular path
4 The speed is non-uniform and the particle is moving on a straight line path
Explanation:
A See above graph. The velocity of the particle is uniform
NDA (I) 2013
Motion in One Dimensions
141391 The plot given above represents displacement ' \(x\) ' of a particle with time ' \(t\) '. The particle is
1 moving with uniform velocity
2 moving with acceleration
3 moving with deceleration
4 executing a periodic motion
Explanation:
B The above given graph is an upward parabola. \(s=u t+\frac{1}{2} a t^{2}\) The above equation of motion is also a parabolic equation. For the upward parabolic equation the \(\mathrm{t}^{2}\) term should be positive. Hence acceleration of the particle is positive. Thus the particle is moving with acceleration. (b) Speed, Velocity and Acceleration
NDA (II) 2013
Motion in One Dimensions
141392
The position \(x\) of a particle varies with time \((t)\) as \(x=\mathbf{A t}^{2}-\mathbf{B} \mathbf{t}^{3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\) ?
1 \(\frac{2 \mathrm{~A}}{3 \mathrm{~B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{\mathrm{A}}{3 \mathrm{~B}}\)
4 zero
Explanation:
C Given that, \(\mathrm{x}=\mathrm{At}^{2}-\mathrm{Bt}^{3}\) \((v)=\frac{d x}{d t}=2 A t-3 B t^{2}\) Acceleration \((\mathrm{a})=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{At}-3 \mathrm{Bt}^{2}\right)\) \(\mathrm{a}=2 \mathrm{~A}-6 \mathrm{Bt}\) Put \(\mathrm{a}=0\) \(2 \mathrm{~A}=6 \mathrm{Bt}\) \(\mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}\) \(\mathrm{t}=\frac{\mathrm{A}}{3 \mathrm{~B}}\)
141389
If \(d\) denotes the distance covered by a car in time \(t\) and \(\vec{S}\) denotes the displacement by the car during the same time, then:
1 \(d \leq|\vec{S}|\)
2 \(d=|\vec{S}|\)
3 \(d \geq|\vec{S}|\)
4 \(d \lt |\vec{S}|\)
Explanation:
C Given, If \(d\) denotes the distance covered by a car in time \(t\) and \(\overrightarrow{\mathrm{S}}\) denotes the displacement by the car during the same that, Displacement is the shortest distance covered by a body and distance is actual path covered by a moving object. Therefore, \(\mathrm{A}-\mathrm{B}-\mathrm{C}\) distance A \(-\mathrm{C}\) displacement \(\mathrm{d} \geq|\overrightarrow{\mathrm{S}}|\)
NDA (I) 2013
Motion in One Dimensions
141390 The motion of a particle is given by a straight line in the graph given above drawn with displacement \((x)\) and time ( \(t\) ). Which one among the following statements is correct?
1 The velocity of the particle is uniform
2 The velocity of the particle is non-uniform
3 The speed is uniform and the particle is moving on a circular path
4 The speed is non-uniform and the particle is moving on a straight line path
Explanation:
A See above graph. The velocity of the particle is uniform
NDA (I) 2013
Motion in One Dimensions
141391 The plot given above represents displacement ' \(x\) ' of a particle with time ' \(t\) '. The particle is
1 moving with uniform velocity
2 moving with acceleration
3 moving with deceleration
4 executing a periodic motion
Explanation:
B The above given graph is an upward parabola. \(s=u t+\frac{1}{2} a t^{2}\) The above equation of motion is also a parabolic equation. For the upward parabolic equation the \(\mathrm{t}^{2}\) term should be positive. Hence acceleration of the particle is positive. Thus the particle is moving with acceleration. (b) Speed, Velocity and Acceleration
NDA (II) 2013
Motion in One Dimensions
141392
The position \(x\) of a particle varies with time \((t)\) as \(x=\mathbf{A t}^{2}-\mathbf{B} \mathbf{t}^{3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\) ?
1 \(\frac{2 \mathrm{~A}}{3 \mathrm{~B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{\mathrm{A}}{3 \mathrm{~B}}\)
4 zero
Explanation:
C Given that, \(\mathrm{x}=\mathrm{At}^{2}-\mathrm{Bt}^{3}\) \((v)=\frac{d x}{d t}=2 A t-3 B t^{2}\) Acceleration \((\mathrm{a})=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{At}-3 \mathrm{Bt}^{2}\right)\) \(\mathrm{a}=2 \mathrm{~A}-6 \mathrm{Bt}\) Put \(\mathrm{a}=0\) \(2 \mathrm{~A}=6 \mathrm{Bt}\) \(\mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}\) \(\mathrm{t}=\frac{\mathrm{A}}{3 \mathrm{~B}}\)
141389
If \(d\) denotes the distance covered by a car in time \(t\) and \(\vec{S}\) denotes the displacement by the car during the same time, then:
1 \(d \leq|\vec{S}|\)
2 \(d=|\vec{S}|\)
3 \(d \geq|\vec{S}|\)
4 \(d \lt |\vec{S}|\)
Explanation:
C Given, If \(d\) denotes the distance covered by a car in time \(t\) and \(\overrightarrow{\mathrm{S}}\) denotes the displacement by the car during the same that, Displacement is the shortest distance covered by a body and distance is actual path covered by a moving object. Therefore, \(\mathrm{A}-\mathrm{B}-\mathrm{C}\) distance A \(-\mathrm{C}\) displacement \(\mathrm{d} \geq|\overrightarrow{\mathrm{S}}|\)
NDA (I) 2013
Motion in One Dimensions
141390 The motion of a particle is given by a straight line in the graph given above drawn with displacement \((x)\) and time ( \(t\) ). Which one among the following statements is correct?
1 The velocity of the particle is uniform
2 The velocity of the particle is non-uniform
3 The speed is uniform and the particle is moving on a circular path
4 The speed is non-uniform and the particle is moving on a straight line path
Explanation:
A See above graph. The velocity of the particle is uniform
NDA (I) 2013
Motion in One Dimensions
141391 The plot given above represents displacement ' \(x\) ' of a particle with time ' \(t\) '. The particle is
1 moving with uniform velocity
2 moving with acceleration
3 moving with deceleration
4 executing a periodic motion
Explanation:
B The above given graph is an upward parabola. \(s=u t+\frac{1}{2} a t^{2}\) The above equation of motion is also a parabolic equation. For the upward parabolic equation the \(\mathrm{t}^{2}\) term should be positive. Hence acceleration of the particle is positive. Thus the particle is moving with acceleration. (b) Speed, Velocity and Acceleration
NDA (II) 2013
Motion in One Dimensions
141392
The position \(x\) of a particle varies with time \((t)\) as \(x=\mathbf{A t}^{2}-\mathbf{B} \mathbf{t}^{3}\). The acceleration at time \(t\) of the particle will be equal to zero. What is the value of \(t\) ?
1 \(\frac{2 \mathrm{~A}}{3 \mathrm{~B}}\)
2 \(\frac{A}{B}\)
3 \(\frac{\mathrm{A}}{3 \mathrm{~B}}\)
4 zero
Explanation:
C Given that, \(\mathrm{x}=\mathrm{At}^{2}-\mathrm{Bt}^{3}\) \((v)=\frac{d x}{d t}=2 A t-3 B t^{2}\) Acceleration \((\mathrm{a})=\frac{\mathrm{d}^{2} \mathrm{x}}{\mathrm{dt}^{2}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(2 \mathrm{At}-3 \mathrm{Bt}^{2}\right)\) \(\mathrm{a}=2 \mathrm{~A}-6 \mathrm{Bt}\) Put \(\mathrm{a}=0\) \(2 \mathrm{~A}=6 \mathrm{Bt}\) \(\mathrm{t}=\frac{2 \mathrm{~A}}{6 \mathrm{~B}}\) \(\mathrm{t}=\frac{\mathrm{A}}{3 \mathrm{~B}}\)
