141154
A particle located at \(x=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=\alpha \sqrt{x}\) where \(\alpha\) is dimensionless constant. The displacement of the particle varies with time as
1 \(t^{3}\)
2 \(t^{2}\)
3 \(\mathrm{t}\)
4 \(t^{1 / 2}\)
Explanation:
B Given, \(\mathrm{v}=\alpha \sqrt{\mathrm{x}} \quad\) (velocity in \(\mathrm{x}\)-direction) Differentiating w.r.t time, \(\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \sqrt{\mathrm{x}}\) \(\int_{0}^{x} \frac{d x}{\sqrt{x}}=\alpha \int_{0}^{t} d t\) \([2 \sqrt{\mathrm{x}}]_{0}^{\mathrm{x}}=\alpha[\mathrm{t}]_{0}^{\mathrm{t}}\) \(x=\frac{\alpha^{2}}{4} t^{2}\) Hence, \(\mathrm{x} \propto \mathrm{t}^{2}\)
AMU-2015
Motion in One Dimensions
141156
A person moves \(30 \mathrm{~m}\) north and then \(20 \mathrm{~m}\) towards east and finally \(30 \sqrt{2} \mathrm{~m}\) in south-west direction. The displacement of the person from the origin will be
1 \(10 \mathrm{~m}\) along north
2 \(10 \mathrm{~m}\) along south
3 \(10 \mathrm{~m}\) along west
4 zero
Explanation:
C The person starts from A- Then, he moves to \(\mathrm{B}\) and then to \(\mathrm{C}\) and finally to \(\mathrm{D}\). In \(\triangle \mathrm{DEC}\), \(\mathrm{CD}^{2}=\mathrm{ED}^{2}+\mathrm{CE}^{2}\) \(\mathrm{ED}^{2}=(30 \sqrt{2})^{2}-(30)^{2}\) \(\mathrm{ED}^{2}=1800-900\) \(\mathrm{ED}^{2}=900\) \(\mathrm{ED}=30 \mathrm{~m}\) \(\therefore\) The displacement of the person from the origin will be \(A D=D E-A E=30-20=10 m\) The displacement of the person from the origin is \(10 \mathrm{~m}\) in the west.
TS- EAMCET-11.09.2020
Motion in One Dimensions
141157
The bus moving with a speed of \(42 \mathrm{~km} / \mathrm{hr}\) is brought to a stop by brakes after \(6 \mathrm{~m}\). If the same bus is moving at a speed of \(90 \mathrm{~km} / \mathrm{hr}\), then the minimum stopping distance is
1 \(15.48 \mathrm{~m}\)
2 \(18.64 \mathrm{~m}\)
3 \(22.13 \mathrm{~m}\)
4 \(27.55 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{u}_{1}=0, \mathrm{~s}_{1}=6 \mathrm{~m}\) \(\mathrm{u}_{2}=0, \mathrm{~s}_{2}=?\) \(\mathrm{u}_{1}=42 \times \frac{5}{18}=11.66 \mathrm{~m} / \mathrm{s}\) \(\mathrm{u}_{2}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\) For first case- \(\mathrm{v}_{1}^{2}=\mathrm{u}_{1}^{2}+2 \mathrm{as}_{1}\) \(0=(11.66)^{2}+2 \mathrm{a} \times 6\) \(\mathrm{a}=\frac{-11.66 \times 11.66}{12}=-11.33 \mathrm{~m} / \mathrm{s}^{2}\) For second case- \(\mathrm{v}_{2}^{2}=\mathrm{u}_{2}^{2}+2 \mathrm{as}_{2}\) \(0=(25)^{2}+2(-11.33) \times \mathrm{s}_{2}\) \(\mathrm{~s}_{2}=\frac{625}{2 \times 11.33}=27.5 \mathrm{~m}\)
MHT-CET-2020
Motion in One Dimensions
141159
A particle starts moving from rest with uniform acceleration. It travels a distance \(x\) in first 2 seconds and distance \(y\) in the next 2 seconds. Then
1 \(y=2 x\)
2 \(y=3 x\)
3 \(y=4 x\)
4 \(y=x\)
Explanation:
B Given that, Let \(\mathrm{x}=\) distance covered in \(2 \mathrm{sec}\) \(\mathrm{x}=\frac{1}{2} \cdot \mathrm{a} \cdot \mathrm{t}^{2}=\frac{1}{2} \cdot \mathrm{a}(2)^{2}=2 \mathrm{a}\) And \(\mathrm{y}=\) distance covered in next \(2 \mathrm{sec}\) \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} \cdot a \cdot(2)^{2}\) \(\mathrm{y}=8 \mathrm{a}-2 \mathrm{a}\) \(\therefore \mathrm{y}=6 \mathrm{a}\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 \mathrm{x}=\mathrm{y}\) \(\therefore \mathrm{y}=3 \mathrm{x}\)
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Motion in One Dimensions
141154
A particle located at \(x=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=\alpha \sqrt{x}\) where \(\alpha\) is dimensionless constant. The displacement of the particle varies with time as
1 \(t^{3}\)
2 \(t^{2}\)
3 \(\mathrm{t}\)
4 \(t^{1 / 2}\)
Explanation:
B Given, \(\mathrm{v}=\alpha \sqrt{\mathrm{x}} \quad\) (velocity in \(\mathrm{x}\)-direction) Differentiating w.r.t time, \(\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \sqrt{\mathrm{x}}\) \(\int_{0}^{x} \frac{d x}{\sqrt{x}}=\alpha \int_{0}^{t} d t\) \([2 \sqrt{\mathrm{x}}]_{0}^{\mathrm{x}}=\alpha[\mathrm{t}]_{0}^{\mathrm{t}}\) \(x=\frac{\alpha^{2}}{4} t^{2}\) Hence, \(\mathrm{x} \propto \mathrm{t}^{2}\)
AMU-2015
Motion in One Dimensions
141156
A person moves \(30 \mathrm{~m}\) north and then \(20 \mathrm{~m}\) towards east and finally \(30 \sqrt{2} \mathrm{~m}\) in south-west direction. The displacement of the person from the origin will be
1 \(10 \mathrm{~m}\) along north
2 \(10 \mathrm{~m}\) along south
3 \(10 \mathrm{~m}\) along west
4 zero
Explanation:
C The person starts from A- Then, he moves to \(\mathrm{B}\) and then to \(\mathrm{C}\) and finally to \(\mathrm{D}\). In \(\triangle \mathrm{DEC}\), \(\mathrm{CD}^{2}=\mathrm{ED}^{2}+\mathrm{CE}^{2}\) \(\mathrm{ED}^{2}=(30 \sqrt{2})^{2}-(30)^{2}\) \(\mathrm{ED}^{2}=1800-900\) \(\mathrm{ED}^{2}=900\) \(\mathrm{ED}=30 \mathrm{~m}\) \(\therefore\) The displacement of the person from the origin will be \(A D=D E-A E=30-20=10 m\) The displacement of the person from the origin is \(10 \mathrm{~m}\) in the west.
TS- EAMCET-11.09.2020
Motion in One Dimensions
141157
The bus moving with a speed of \(42 \mathrm{~km} / \mathrm{hr}\) is brought to a stop by brakes after \(6 \mathrm{~m}\). If the same bus is moving at a speed of \(90 \mathrm{~km} / \mathrm{hr}\), then the minimum stopping distance is
1 \(15.48 \mathrm{~m}\)
2 \(18.64 \mathrm{~m}\)
3 \(22.13 \mathrm{~m}\)
4 \(27.55 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{u}_{1}=0, \mathrm{~s}_{1}=6 \mathrm{~m}\) \(\mathrm{u}_{2}=0, \mathrm{~s}_{2}=?\) \(\mathrm{u}_{1}=42 \times \frac{5}{18}=11.66 \mathrm{~m} / \mathrm{s}\) \(\mathrm{u}_{2}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\) For first case- \(\mathrm{v}_{1}^{2}=\mathrm{u}_{1}^{2}+2 \mathrm{as}_{1}\) \(0=(11.66)^{2}+2 \mathrm{a} \times 6\) \(\mathrm{a}=\frac{-11.66 \times 11.66}{12}=-11.33 \mathrm{~m} / \mathrm{s}^{2}\) For second case- \(\mathrm{v}_{2}^{2}=\mathrm{u}_{2}^{2}+2 \mathrm{as}_{2}\) \(0=(25)^{2}+2(-11.33) \times \mathrm{s}_{2}\) \(\mathrm{~s}_{2}=\frac{625}{2 \times 11.33}=27.5 \mathrm{~m}\)
MHT-CET-2020
Motion in One Dimensions
141159
A particle starts moving from rest with uniform acceleration. It travels a distance \(x\) in first 2 seconds and distance \(y\) in the next 2 seconds. Then
1 \(y=2 x\)
2 \(y=3 x\)
3 \(y=4 x\)
4 \(y=x\)
Explanation:
B Given that, Let \(\mathrm{x}=\) distance covered in \(2 \mathrm{sec}\) \(\mathrm{x}=\frac{1}{2} \cdot \mathrm{a} \cdot \mathrm{t}^{2}=\frac{1}{2} \cdot \mathrm{a}(2)^{2}=2 \mathrm{a}\) And \(\mathrm{y}=\) distance covered in next \(2 \mathrm{sec}\) \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} \cdot a \cdot(2)^{2}\) \(\mathrm{y}=8 \mathrm{a}-2 \mathrm{a}\) \(\therefore \mathrm{y}=6 \mathrm{a}\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 \mathrm{x}=\mathrm{y}\) \(\therefore \mathrm{y}=3 \mathrm{x}\)
141154
A particle located at \(x=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=\alpha \sqrt{x}\) where \(\alpha\) is dimensionless constant. The displacement of the particle varies with time as
1 \(t^{3}\)
2 \(t^{2}\)
3 \(\mathrm{t}\)
4 \(t^{1 / 2}\)
Explanation:
B Given, \(\mathrm{v}=\alpha \sqrt{\mathrm{x}} \quad\) (velocity in \(\mathrm{x}\)-direction) Differentiating w.r.t time, \(\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \sqrt{\mathrm{x}}\) \(\int_{0}^{x} \frac{d x}{\sqrt{x}}=\alpha \int_{0}^{t} d t\) \([2 \sqrt{\mathrm{x}}]_{0}^{\mathrm{x}}=\alpha[\mathrm{t}]_{0}^{\mathrm{t}}\) \(x=\frac{\alpha^{2}}{4} t^{2}\) Hence, \(\mathrm{x} \propto \mathrm{t}^{2}\)
AMU-2015
Motion in One Dimensions
141156
A person moves \(30 \mathrm{~m}\) north and then \(20 \mathrm{~m}\) towards east and finally \(30 \sqrt{2} \mathrm{~m}\) in south-west direction. The displacement of the person from the origin will be
1 \(10 \mathrm{~m}\) along north
2 \(10 \mathrm{~m}\) along south
3 \(10 \mathrm{~m}\) along west
4 zero
Explanation:
C The person starts from A- Then, he moves to \(\mathrm{B}\) and then to \(\mathrm{C}\) and finally to \(\mathrm{D}\). In \(\triangle \mathrm{DEC}\), \(\mathrm{CD}^{2}=\mathrm{ED}^{2}+\mathrm{CE}^{2}\) \(\mathrm{ED}^{2}=(30 \sqrt{2})^{2}-(30)^{2}\) \(\mathrm{ED}^{2}=1800-900\) \(\mathrm{ED}^{2}=900\) \(\mathrm{ED}=30 \mathrm{~m}\) \(\therefore\) The displacement of the person from the origin will be \(A D=D E-A E=30-20=10 m\) The displacement of the person from the origin is \(10 \mathrm{~m}\) in the west.
TS- EAMCET-11.09.2020
Motion in One Dimensions
141157
The bus moving with a speed of \(42 \mathrm{~km} / \mathrm{hr}\) is brought to a stop by brakes after \(6 \mathrm{~m}\). If the same bus is moving at a speed of \(90 \mathrm{~km} / \mathrm{hr}\), then the minimum stopping distance is
1 \(15.48 \mathrm{~m}\)
2 \(18.64 \mathrm{~m}\)
3 \(22.13 \mathrm{~m}\)
4 \(27.55 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{u}_{1}=0, \mathrm{~s}_{1}=6 \mathrm{~m}\) \(\mathrm{u}_{2}=0, \mathrm{~s}_{2}=?\) \(\mathrm{u}_{1}=42 \times \frac{5}{18}=11.66 \mathrm{~m} / \mathrm{s}\) \(\mathrm{u}_{2}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\) For first case- \(\mathrm{v}_{1}^{2}=\mathrm{u}_{1}^{2}+2 \mathrm{as}_{1}\) \(0=(11.66)^{2}+2 \mathrm{a} \times 6\) \(\mathrm{a}=\frac{-11.66 \times 11.66}{12}=-11.33 \mathrm{~m} / \mathrm{s}^{2}\) For second case- \(\mathrm{v}_{2}^{2}=\mathrm{u}_{2}^{2}+2 \mathrm{as}_{2}\) \(0=(25)^{2}+2(-11.33) \times \mathrm{s}_{2}\) \(\mathrm{~s}_{2}=\frac{625}{2 \times 11.33}=27.5 \mathrm{~m}\)
MHT-CET-2020
Motion in One Dimensions
141159
A particle starts moving from rest with uniform acceleration. It travels a distance \(x\) in first 2 seconds and distance \(y\) in the next 2 seconds. Then
1 \(y=2 x\)
2 \(y=3 x\)
3 \(y=4 x\)
4 \(y=x\)
Explanation:
B Given that, Let \(\mathrm{x}=\) distance covered in \(2 \mathrm{sec}\) \(\mathrm{x}=\frac{1}{2} \cdot \mathrm{a} \cdot \mathrm{t}^{2}=\frac{1}{2} \cdot \mathrm{a}(2)^{2}=2 \mathrm{a}\) And \(\mathrm{y}=\) distance covered in next \(2 \mathrm{sec}\) \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} \cdot a \cdot(2)^{2}\) \(\mathrm{y}=8 \mathrm{a}-2 \mathrm{a}\) \(\therefore \mathrm{y}=6 \mathrm{a}\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 \mathrm{x}=\mathrm{y}\) \(\therefore \mathrm{y}=3 \mathrm{x}\)
141154
A particle located at \(x=0\), starts moving along the positive \(x\)-direction with a velocity \(v\) that varies as \(v=\alpha \sqrt{x}\) where \(\alpha\) is dimensionless constant. The displacement of the particle varies with time as
1 \(t^{3}\)
2 \(t^{2}\)
3 \(\mathrm{t}\)
4 \(t^{1 / 2}\)
Explanation:
B Given, \(\mathrm{v}=\alpha \sqrt{\mathrm{x}} \quad\) (velocity in \(\mathrm{x}\)-direction) Differentiating w.r.t time, \(\frac{\mathrm{dx}}{\mathrm{dt}}=\alpha \sqrt{\mathrm{x}}\) \(\int_{0}^{x} \frac{d x}{\sqrt{x}}=\alpha \int_{0}^{t} d t\) \([2 \sqrt{\mathrm{x}}]_{0}^{\mathrm{x}}=\alpha[\mathrm{t}]_{0}^{\mathrm{t}}\) \(x=\frac{\alpha^{2}}{4} t^{2}\) Hence, \(\mathrm{x} \propto \mathrm{t}^{2}\)
AMU-2015
Motion in One Dimensions
141156
A person moves \(30 \mathrm{~m}\) north and then \(20 \mathrm{~m}\) towards east and finally \(30 \sqrt{2} \mathrm{~m}\) in south-west direction. The displacement of the person from the origin will be
1 \(10 \mathrm{~m}\) along north
2 \(10 \mathrm{~m}\) along south
3 \(10 \mathrm{~m}\) along west
4 zero
Explanation:
C The person starts from A- Then, he moves to \(\mathrm{B}\) and then to \(\mathrm{C}\) and finally to \(\mathrm{D}\). In \(\triangle \mathrm{DEC}\), \(\mathrm{CD}^{2}=\mathrm{ED}^{2}+\mathrm{CE}^{2}\) \(\mathrm{ED}^{2}=(30 \sqrt{2})^{2}-(30)^{2}\) \(\mathrm{ED}^{2}=1800-900\) \(\mathrm{ED}^{2}=900\) \(\mathrm{ED}=30 \mathrm{~m}\) \(\therefore\) The displacement of the person from the origin will be \(A D=D E-A E=30-20=10 m\) The displacement of the person from the origin is \(10 \mathrm{~m}\) in the west.
TS- EAMCET-11.09.2020
Motion in One Dimensions
141157
The bus moving with a speed of \(42 \mathrm{~km} / \mathrm{hr}\) is brought to a stop by brakes after \(6 \mathrm{~m}\). If the same bus is moving at a speed of \(90 \mathrm{~km} / \mathrm{hr}\), then the minimum stopping distance is
1 \(15.48 \mathrm{~m}\)
2 \(18.64 \mathrm{~m}\)
3 \(22.13 \mathrm{~m}\)
4 \(27.55 \mathrm{~m}\)
Explanation:
D Given, \(\mathrm{u}_{1}=0, \mathrm{~s}_{1}=6 \mathrm{~m}\) \(\mathrm{u}_{2}=0, \mathrm{~s}_{2}=?\) \(\mathrm{u}_{1}=42 \times \frac{5}{18}=11.66 \mathrm{~m} / \mathrm{s}\) \(\mathrm{u}_{2}=90 \times \frac{5}{18}=25 \mathrm{~m} / \mathrm{s}\) For first case- \(\mathrm{v}_{1}^{2}=\mathrm{u}_{1}^{2}+2 \mathrm{as}_{1}\) \(0=(11.66)^{2}+2 \mathrm{a} \times 6\) \(\mathrm{a}=\frac{-11.66 \times 11.66}{12}=-11.33 \mathrm{~m} / \mathrm{s}^{2}\) For second case- \(\mathrm{v}_{2}^{2}=\mathrm{u}_{2}^{2}+2 \mathrm{as}_{2}\) \(0=(25)^{2}+2(-11.33) \times \mathrm{s}_{2}\) \(\mathrm{~s}_{2}=\frac{625}{2 \times 11.33}=27.5 \mathrm{~m}\)
MHT-CET-2020
Motion in One Dimensions
141159
A particle starts moving from rest with uniform acceleration. It travels a distance \(x\) in first 2 seconds and distance \(y\) in the next 2 seconds. Then
1 \(y=2 x\)
2 \(y=3 x\)
3 \(y=4 x\)
4 \(y=x\)
Explanation:
B Given that, Let \(\mathrm{x}=\) distance covered in \(2 \mathrm{sec}\) \(\mathrm{x}=\frac{1}{2} \cdot \mathrm{a} \cdot \mathrm{t}^{2}=\frac{1}{2} \cdot \mathrm{a}(2)^{2}=2 \mathrm{a}\) And \(\mathrm{y}=\) distance covered in next \(2 \mathrm{sec}\) \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} \cdot a \cdot(2)^{2}\) \(\mathrm{y}=8 \mathrm{a}-2 \mathrm{a}\) \(\therefore \mathrm{y}=6 \mathrm{a}\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 \mathrm{x}=\mathrm{y}\) \(\therefore \mathrm{y}=3 \mathrm{x}\)
