QBManager

Motion in One Dimensions

141250 The equation of motion of a particle is given by \(x(t)=2 t^{3}+t^{2}+4 t\). Then the average velocity of the particle between the time \(t=3 \mathrm{~s}\) to \(t=5 \mathrm{~s}\) is

1 \(100 \mathrm{~m} / \mathrm{s}\)

2 \(110 \mathrm{~m} / \mathrm{s}\)

3 \(90 \mathrm{~m} / \mathrm{s}\)

4 \(140 \mathrm{~m} / \mathrm{s}\)

TS EAMCET 02.05.2018

Motion in One Dimensions

141251 The displacement of a body is given by \(x=4 t+\) \(5 t^{3}\), where \(x\) is in meter and \(t\) is in second. The difference between the average velocity of the body in the time-interval \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\) and its instantaneous velocity at \(t=1 \mathrm{~s}\) is

1 \(20.0 \mathrm{~m} \mathrm{~s}^{-1}\)

2 \(22.5 \mathrm{~m} \mathrm{~s}^{-1}\)

3 \(27.0 \mathrm{~m} \mathrm{~s}^{-1}\)

4 \(39.0 \mathrm{~m} \mathrm{~s}^{-1}\)

AMU-2018

Motion in One Dimensions

141252 The nearest star to our solar system is 4.3 light years away. The distance of this star in Parsec is (Mean distance between the earth and the sun \(=1.5 \times 10^{11} \mathrm{~m}\) and one light year \(=9.46 \times\) \(\left.10^{15} \mathrm{~m}\right)\)

1 1.3

2 8.0

3 13.0

4 \(3.3 \times 10^{4}\)

AMU-2018

Motion in One Dimensions

141253 The velocity displacement ( \(\mathrm{v}\)-s) graph shows the motion of particle moving in a straight line. Velocity-displacement graph is a circle of radius \(2 \mathrm{~m}\) and centre is at \((2,0) \mathrm{m}\).
original image
The value of acceleration for this particle at a point \((2-\sqrt{2}, \sqrt{2}) \mathrm{m}\) will be \(\mathrm{ms}^{-2}\).

1 \(\sqrt{2}\)

2 4

3 2

4 3

AP EAMCET (23.04.2018) Shift-2

Motion in One Dimensions

141254 A particle is moving along the \(\mathrm{X}\)-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement \(x(t)\) is given by

1 \(\sin \omega t, \omega>0\)

2 \(\sin \omega t+\cos \omega t, \omega>0\)

3 \(\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)

4 \(\mathrm{e}^{\omega t}+\sin \omega \mathrm{t}, \omega>0\)

5 \(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}} \omega_{1}\) and \(\omega_{2}>0\)

Explanation:

C Given,
displacement \(=x(t)\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t})\)
According to option (a),
\(\mathrm{x}(\mathrm{t})=\sin \omega \mathrm{t}, \omega>0\)
\(u(t)=\frac{d x(t)}{d t}=\frac{d}{d t} \sin \omega t=\omega \cos \omega t \text { and }\)
\(a=\frac{d u(t)}{d t}=\frac{d}{d t}(\omega \cos \omega t)=-\omega^{2} \sin \omega t\)
\(\text { Or } \quad a=-\omega^{2} x(t)\)
\(\text { in the opposite direction }\)
\(\text { According to option }(b),\)
\(x(t)=\sin \omega t+\cos \omega t, \omega>0\)
\(\therefore \quad u(t)=\frac{d x(t)}{d t}=\omega \cos \omega t-\omega \sin \omega t\)
\(\text { and } \quad a=\frac{d u(t)}{d t}=-\omega^{2} \sin \omega t-\omega^{2} \cos \omega t\)
\(\text { or } a=-\left(\omega^{2} \sin \omega t+\omega^{2} \cos \omega t\right) \text { or } a=-\omega^{2} x(t)\)
Thus, It also in opposite direction.
According to option (c),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{x}(\mathrm{t})\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t}) \text { in the same direction }\)
According to option (d),
\(\therefore \quad \mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\sin \omega \mathrm{t} \quad \omega>0\)
\(\therefore \quad \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+\omega \cos \omega \mathrm{t}\)
\(\text { and } \mathrm{a}= \frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}-\omega^{2} \sin \omega \mathrm{t}\)
\(\text { or } \quad \mathrm{a}=\omega^{2}\left(\mathrm{e}^{\omega \mathrm{t}}-\sin \omega \mathrm{t}\right)\)
According to option (e),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}}, \omega_{1} \text { and } \omega_{2}>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+2 \omega \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
Hence, by comparing equation (ii), (iii), (iv), (v) and (vi) with Eq. (i) option (c) is best suited.

Kerala CEE -2018

Motion in One Dimensions

141250 The equation of motion of a particle is given by \(x(t)=2 t^{3}+t^{2}+4 t\). Then the average velocity of the particle between the time \(t=3 \mathrm{~s}\) to \(t=5 \mathrm{~s}\) is

1 \(100 \mathrm{~m} / \mathrm{s}\)

2 \(110 \mathrm{~m} / \mathrm{s}\)

3 \(90 \mathrm{~m} / \mathrm{s}\)

4 \(140 \mathrm{~m} / \mathrm{s}\)

TS EAMCET 02.05.2018

Motion in One Dimensions

141251 The displacement of a body is given by \(x=4 t+\) \(5 t^{3}\), where \(x\) is in meter and \(t\) is in second. The difference between the average velocity of the body in the time-interval \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\) and its instantaneous velocity at \(t=1 \mathrm{~s}\) is

1 \(20.0 \mathrm{~m} \mathrm{~s}^{-1}\)

2 \(22.5 \mathrm{~m} \mathrm{~s}^{-1}\)

3 \(27.0 \mathrm{~m} \mathrm{~s}^{-1}\)

4 \(39.0 \mathrm{~m} \mathrm{~s}^{-1}\)

AMU-2018

Motion in One Dimensions

141252 The nearest star to our solar system is 4.3 light years away. The distance of this star in Parsec is (Mean distance between the earth and the sun \(=1.5 \times 10^{11} \mathrm{~m}\) and one light year \(=9.46 \times\) \(\left.10^{15} \mathrm{~m}\right)\)

1 1.3

2 8.0

3 13.0

4 \(3.3 \times 10^{4}\)

AMU-2018

Motion in One Dimensions

141253 The velocity displacement ( \(\mathrm{v}\)-s) graph shows the motion of particle moving in a straight line. Velocity-displacement graph is a circle of radius \(2 \mathrm{~m}\) and centre is at \((2,0) \mathrm{m}\).
original image
The value of acceleration for this particle at a point \((2-\sqrt{2}, \sqrt{2}) \mathrm{m}\) will be \(\mathrm{ms}^{-2}\).

1 \(\sqrt{2}\)

2 4

3 2

4 3

AP EAMCET (23.04.2018) Shift-2

Motion in One Dimensions

141254 A particle is moving along the \(\mathrm{X}\)-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement \(x(t)\) is given by

1 \(\sin \omega t, \omega>0\)

2 \(\sin \omega t+\cos \omega t, \omega>0\)

3 \(\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)

4 \(\mathrm{e}^{\omega t}+\sin \omega \mathrm{t}, \omega>0\)

5 \(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}} \omega_{1}\) and \(\omega_{2}>0\)

Explanation:

C Given,
displacement \(=x(t)\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t})\)
According to option (a),
\(\mathrm{x}(\mathrm{t})=\sin \omega \mathrm{t}, \omega>0\)
\(u(t)=\frac{d x(t)}{d t}=\frac{d}{d t} \sin \omega t=\omega \cos \omega t \text { and }\)
\(a=\frac{d u(t)}{d t}=\frac{d}{d t}(\omega \cos \omega t)=-\omega^{2} \sin \omega t\)
\(\text { Or } \quad a=-\omega^{2} x(t)\)
\(\text { in the opposite direction }\)
\(\text { According to option }(b),\)
\(x(t)=\sin \omega t+\cos \omega t, \omega>0\)
\(\therefore \quad u(t)=\frac{d x(t)}{d t}=\omega \cos \omega t-\omega \sin \omega t\)
\(\text { and } \quad a=\frac{d u(t)}{d t}=-\omega^{2} \sin \omega t-\omega^{2} \cos \omega t\)
\(\text { or } a=-\left(\omega^{2} \sin \omega t+\omega^{2} \cos \omega t\right) \text { or } a=-\omega^{2} x(t)\)
Thus, It also in opposite direction.
According to option (c),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{x}(\mathrm{t})\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t}) \text { in the same direction }\)
According to option (d),
\(\therefore \quad \mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\sin \omega \mathrm{t} \quad \omega>0\)
\(\therefore \quad \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+\omega \cos \omega \mathrm{t}\)
\(\text { and } \mathrm{a}= \frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}-\omega^{2} \sin \omega \mathrm{t}\)
\(\text { or } \quad \mathrm{a}=\omega^{2}\left(\mathrm{e}^{\omega \mathrm{t}}-\sin \omega \mathrm{t}\right)\)
According to option (e),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}}, \omega_{1} \text { and } \omega_{2}>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+2 \omega \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
Hence, by comparing equation (ii), (iii), (iv), (v) and (vi) with Eq. (i) option (c) is best suited.

Kerala CEE -2018

Motion in One Dimensions

141250 The equation of motion of a particle is given by \(x(t)=2 t^{3}+t^{2}+4 t\). Then the average velocity of the particle between the time \(t=3 \mathrm{~s}\) to \(t=5 \mathrm{~s}\) is

1 \(100 \mathrm{~m} / \mathrm{s}\)

2 \(110 \mathrm{~m} / \mathrm{s}\)

3 \(90 \mathrm{~m} / \mathrm{s}\)

4 \(140 \mathrm{~m} / \mathrm{s}\)

TS EAMCET 02.05.2018

Motion in One Dimensions

141251 The displacement of a body is given by \(x=4 t+\) \(5 t^{3}\), where \(x\) is in meter and \(t\) is in second. The difference between the average velocity of the body in the time-interval \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\) and its instantaneous velocity at \(t=1 \mathrm{~s}\) is

1 \(20.0 \mathrm{~m} \mathrm{~s}^{-1}\)

2 \(22.5 \mathrm{~m} \mathrm{~s}^{-1}\)

3 \(27.0 \mathrm{~m} \mathrm{~s}^{-1}\)

4 \(39.0 \mathrm{~m} \mathrm{~s}^{-1}\)

AMU-2018

Motion in One Dimensions

141252 The nearest star to our solar system is 4.3 light years away. The distance of this star in Parsec is (Mean distance between the earth and the sun \(=1.5 \times 10^{11} \mathrm{~m}\) and one light year \(=9.46 \times\) \(\left.10^{15} \mathrm{~m}\right)\)

1 1.3

2 8.0

3 13.0

4 \(3.3 \times 10^{4}\)

AMU-2018

Motion in One Dimensions

141253 The velocity displacement ( \(\mathrm{v}\)-s) graph shows the motion of particle moving in a straight line. Velocity-displacement graph is a circle of radius \(2 \mathrm{~m}\) and centre is at \((2,0) \mathrm{m}\).
original image
The value of acceleration for this particle at a point \((2-\sqrt{2}, \sqrt{2}) \mathrm{m}\) will be \(\mathrm{ms}^{-2}\).

1 \(\sqrt{2}\)

2 4

3 2

4 3

AP EAMCET (23.04.2018) Shift-2

Motion in One Dimensions

141254 A particle is moving along the \(\mathrm{X}\)-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement \(x(t)\) is given by

1 \(\sin \omega t, \omega>0\)

2 \(\sin \omega t+\cos \omega t, \omega>0\)

3 \(\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)

4 \(\mathrm{e}^{\omega t}+\sin \omega \mathrm{t}, \omega>0\)

5 \(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}} \omega_{1}\) and \(\omega_{2}>0\)

Explanation:

C Given,
displacement \(=x(t)\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t})\)
According to option (a),
\(\mathrm{x}(\mathrm{t})=\sin \omega \mathrm{t}, \omega>0\)
\(u(t)=\frac{d x(t)}{d t}=\frac{d}{d t} \sin \omega t=\omega \cos \omega t \text { and }\)
\(a=\frac{d u(t)}{d t}=\frac{d}{d t}(\omega \cos \omega t)=-\omega^{2} \sin \omega t\)
\(\text { Or } \quad a=-\omega^{2} x(t)\)
\(\text { in the opposite direction }\)
\(\text { According to option }(b),\)
\(x(t)=\sin \omega t+\cos \omega t, \omega>0\)
\(\therefore \quad u(t)=\frac{d x(t)}{d t}=\omega \cos \omega t-\omega \sin \omega t\)
\(\text { and } \quad a=\frac{d u(t)}{d t}=-\omega^{2} \sin \omega t-\omega^{2} \cos \omega t\)
\(\text { or } a=-\left(\omega^{2} \sin \omega t+\omega^{2} \cos \omega t\right) \text { or } a=-\omega^{2} x(t)\)
Thus, It also in opposite direction.
According to option (c),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{x}(\mathrm{t})\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t}) \text { in the same direction }\)
According to option (d),
\(\therefore \quad \mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\sin \omega \mathrm{t} \quad \omega>0\)
\(\therefore \quad \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+\omega \cos \omega \mathrm{t}\)
\(\text { and } \mathrm{a}= \frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}-\omega^{2} \sin \omega \mathrm{t}\)
\(\text { or } \quad \mathrm{a}=\omega^{2}\left(\mathrm{e}^{\omega \mathrm{t}}-\sin \omega \mathrm{t}\right)\)
According to option (e),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}}, \omega_{1} \text { and } \omega_{2}>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+2 \omega \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
Hence, by comparing equation (ii), (iii), (iv), (v) and (vi) with Eq. (i) option (c) is best suited.

Kerala CEE -2018

Motion in One Dimensions

141250 The equation of motion of a particle is given by \(x(t)=2 t^{3}+t^{2}+4 t\). Then the average velocity of the particle between the time \(t=3 \mathrm{~s}\) to \(t=5 \mathrm{~s}\) is

1 \(100 \mathrm{~m} / \mathrm{s}\)

2 \(110 \mathrm{~m} / \mathrm{s}\)

3 \(90 \mathrm{~m} / \mathrm{s}\)

4 \(140 \mathrm{~m} / \mathrm{s}\)

TS EAMCET 02.05.2018

Motion in One Dimensions

141251 The displacement of a body is given by \(x=4 t+\) \(5 t^{3}\), where \(x\) is in meter and \(t\) is in second. The difference between the average velocity of the body in the time-interval \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\) and its instantaneous velocity at \(t=1 \mathrm{~s}\) is

1 \(20.0 \mathrm{~m} \mathrm{~s}^{-1}\)

2 \(22.5 \mathrm{~m} \mathrm{~s}^{-1}\)

3 \(27.0 \mathrm{~m} \mathrm{~s}^{-1}\)

4 \(39.0 \mathrm{~m} \mathrm{~s}^{-1}\)

AMU-2018

Motion in One Dimensions

141252 The nearest star to our solar system is 4.3 light years away. The distance of this star in Parsec is (Mean distance between the earth and the sun \(=1.5 \times 10^{11} \mathrm{~m}\) and one light year \(=9.46 \times\) \(\left.10^{15} \mathrm{~m}\right)\)

1 1.3

2 8.0

3 13.0

4 \(3.3 \times 10^{4}\)

AMU-2018

Motion in One Dimensions

141253 The velocity displacement ( \(\mathrm{v}\)-s) graph shows the motion of particle moving in a straight line. Velocity-displacement graph is a circle of radius \(2 \mathrm{~m}\) and centre is at \((2,0) \mathrm{m}\).
original image
The value of acceleration for this particle at a point \((2-\sqrt{2}, \sqrt{2}) \mathrm{m}\) will be \(\mathrm{ms}^{-2}\).

1 \(\sqrt{2}\)

2 4

3 2

4 3

AP EAMCET (23.04.2018) Shift-2

Motion in One Dimensions

141254 A particle is moving along the \(\mathrm{X}\)-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement \(x(t)\) is given by

1 \(\sin \omega t, \omega>0\)

2 \(\sin \omega t+\cos \omega t, \omega>0\)

3 \(\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)

4 \(\mathrm{e}^{\omega t}+\sin \omega \mathrm{t}, \omega>0\)

5 \(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}} \omega_{1}\) and \(\omega_{2}>0\)

Explanation:

C Given,
displacement \(=x(t)\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t})\)
According to option (a),
\(\mathrm{x}(\mathrm{t})=\sin \omega \mathrm{t}, \omega>0\)
\(u(t)=\frac{d x(t)}{d t}=\frac{d}{d t} \sin \omega t=\omega \cos \omega t \text { and }\)
\(a=\frac{d u(t)}{d t}=\frac{d}{d t}(\omega \cos \omega t)=-\omega^{2} \sin \omega t\)
\(\text { Or } \quad a=-\omega^{2} x(t)\)
\(\text { in the opposite direction }\)
\(\text { According to option }(b),\)
\(x(t)=\sin \omega t+\cos \omega t, \omega>0\)
\(\therefore \quad u(t)=\frac{d x(t)}{d t}=\omega \cos \omega t-\omega \sin \omega t\)
\(\text { and } \quad a=\frac{d u(t)}{d t}=-\omega^{2} \sin \omega t-\omega^{2} \cos \omega t\)
\(\text { or } a=-\left(\omega^{2} \sin \omega t+\omega^{2} \cos \omega t\right) \text { or } a=-\omega^{2} x(t)\)
Thus, It also in opposite direction.
According to option (c),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{x}(\mathrm{t})\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t}) \text { in the same direction }\)
According to option (d),
\(\therefore \quad \mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\sin \omega \mathrm{t} \quad \omega>0\)
\(\therefore \quad \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+\omega \cos \omega \mathrm{t}\)
\(\text { and } \mathrm{a}= \frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}-\omega^{2} \sin \omega \mathrm{t}\)
\(\text { or } \quad \mathrm{a}=\omega^{2}\left(\mathrm{e}^{\omega \mathrm{t}}-\sin \omega \mathrm{t}\right)\)
According to option (e),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}}, \omega_{1} \text { and } \omega_{2}>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+2 \omega \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
Hence, by comparing equation (ii), (iii), (iv), (v) and (vi) with Eq. (i) option (c) is best suited.

Kerala CEE -2018

Motion in One Dimensions

141250 The equation of motion of a particle is given by \(x(t)=2 t^{3}+t^{2}+4 t\). Then the average velocity of the particle between the time \(t=3 \mathrm{~s}\) to \(t=5 \mathrm{~s}\) is

1 \(100 \mathrm{~m} / \mathrm{s}\)

2 \(110 \mathrm{~m} / \mathrm{s}\)

3 \(90 \mathrm{~m} / \mathrm{s}\)

4 \(140 \mathrm{~m} / \mathrm{s}\)

TS EAMCET 02.05.2018

Motion in One Dimensions

141251 The displacement of a body is given by \(x=4 t+\) \(5 t^{3}\), where \(x\) is in meter and \(t\) is in second. The difference between the average velocity of the body in the time-interval \(t=1 \mathrm{~s}\) to \(t=2 \mathrm{~s}\) and its instantaneous velocity at \(t=1 \mathrm{~s}\) is

1 \(20.0 \mathrm{~m} \mathrm{~s}^{-1}\)

2 \(22.5 \mathrm{~m} \mathrm{~s}^{-1}\)

3 \(27.0 \mathrm{~m} \mathrm{~s}^{-1}\)

4 \(39.0 \mathrm{~m} \mathrm{~s}^{-1}\)

AMU-2018

Motion in One Dimensions

141252 The nearest star to our solar system is 4.3 light years away. The distance of this star in Parsec is (Mean distance between the earth and the sun \(=1.5 \times 10^{11} \mathrm{~m}\) and one light year \(=9.46 \times\) \(\left.10^{15} \mathrm{~m}\right)\)

1 1.3

2 8.0

3 13.0

4 \(3.3 \times 10^{4}\)

AMU-2018

Motion in One Dimensions

141253 The velocity displacement ( \(\mathrm{v}\)-s) graph shows the motion of particle moving in a straight line. Velocity-displacement graph is a circle of radius \(2 \mathrm{~m}\) and centre is at \((2,0) \mathrm{m}\).
original image
The value of acceleration for this particle at a point \((2-\sqrt{2}, \sqrt{2}) \mathrm{m}\) will be \(\mathrm{ms}^{-2}\).

1 \(\sqrt{2}\)

2 4

3 2

4 3

AP EAMCET (23.04.2018) Shift-2

Motion in One Dimensions

141254 A particle is moving along the \(\mathrm{X}\)-axis such that its acceleration is proportional to the displacement from the equilibrium position and they are in the same direction. The displacement \(x(t)\) is given by

1 \(\sin \omega t, \omega>0\)

2 \(\sin \omega t+\cos \omega t, \omega>0\)

3 \(\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)

4 \(\mathrm{e}^{\omega t}+\sin \omega \mathrm{t}, \omega>0\)

5 \(\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}} \omega_{1}\) and \(\omega_{2}>0\)

Explanation:

C Given,
displacement \(=x(t)\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t})\)
According to option (a),
\(\mathrm{x}(\mathrm{t})=\sin \omega \mathrm{t}, \omega>0\)
\(u(t)=\frac{d x(t)}{d t}=\frac{d}{d t} \sin \omega t=\omega \cos \omega t \text { and }\)
\(a=\frac{d u(t)}{d t}=\frac{d}{d t}(\omega \cos \omega t)=-\omega^{2} \sin \omega t\)
\(\text { Or } \quad a=-\omega^{2} x(t)\)
\(\text { in the opposite direction }\)
\(\text { According to option }(b),\)
\(x(t)=\sin \omega t+\cos \omega t, \omega>0\)
\(\therefore \quad u(t)=\frac{d x(t)}{d t}=\omega \cos \omega t-\omega \sin \omega t\)
\(\text { and } \quad a=\frac{d u(t)}{d t}=-\omega^{2} \sin \omega t-\omega^{2} \cos \omega t\)
\(\text { or } a=-\left(\omega^{2} \sin \omega t+\omega^{2} \cos \omega t\right) \text { or } a=-\omega^{2} x(t)\)
Thus, It also in opposite direction.
According to option (c),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}, \omega>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{x}(\mathrm{t})\)
\(\mathrm{a} \propto \mathrm{x}(\mathrm{t}) \text { in the same direction }\)
According to option (d),
\(\therefore \quad \mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\sin \omega \mathrm{t} \quad \omega>0\)
\(\therefore \quad \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+\omega \cos \omega \mathrm{t}\)
\(\text { and } \mathrm{a}= \frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}-\omega^{2} \sin \omega \mathrm{t}\)
\(\text { or } \quad \mathrm{a}=\omega^{2}\left(\mathrm{e}^{\omega \mathrm{t}}-\sin \omega \mathrm{t}\right)\)
According to option (e),
\(\mathrm{x}(\mathrm{t})=\mathrm{e}^{\omega \mathrm{t}}+\mathrm{e}^{\omega^{2} \mathrm{t}}, \omega_{1} \text { and } \omega_{2}>0\)
\(\therefore \mathrm{u}(\mathrm{t})=\frac{\mathrm{dx}(\mathrm{t})}{\mathrm{dt}}=\omega \mathrm{e}^{\omega \mathrm{t}}+2 \omega \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { and } \mathrm{a}=\frac{\mathrm{du}(\mathrm{t})}{\mathrm{dt}}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
\(\text { or } \mathrm{a}=\omega^{2} \mathrm{e}^{\omega \mathrm{t}}+4 \omega^{2} \mathrm{e}^{\omega^{2} \mathrm{t}}\)
Hence, by comparing equation (ii), (iii), (iv), (v) and (vi) with Eq. (i) option (c) is best suited.

Kerala CEE -2018

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