141429
To reach point-B from point-A, a person travels \(500 \mathrm{~m}\) North, \(400 \mathrm{~m}\) East and \(200 \mathrm{~m}\) south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person
A Distance \(=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}\) \(=(500+400+200)\) \(=1100 \mathrm{~m}\) Displacement \(=\mathrm{AD}=\sqrt{(\mathrm{AB}-\mathrm{CD})^{2}+\mathrm{BC}^{2}}\) \(=\sqrt{(500-200)^{2}+400^{2}}\) \(=500 \mathrm{~m}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(\qquad \frac{1100}{1200}=\frac{11}{12} \mathrm{~m} / \mathrm{s}\) \(\text { Average Velocity }=\frac{A D}{t}\) \(=\frac{500}{1200}\) Average velocity \(=\frac{5}{12} \mathrm{~ms}^{-1}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141430
A particle is at \(x=0\) when \(t=0\). It moves among \(x\)-axis with a velocity given by \(v=5 \sqrt{x}\). Find the acceleration of this particle
1 \(8.5 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(12.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(10.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(11.5 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
B We know that Acceleration \(=\) derivative of speed with respect to time Therefore, \(a=\frac{d v}{d t}\) \(a=\frac{d v}{d x} \times \frac{d x}{d t}\) \(a=d(5 \sqrt{x}) / d x \times \frac{d x}{d t}\) \(a=\frac{5}{2 \sqrt{x}} \times v \quad(v=5 \sqrt{x})\) \(a=\left(\frac{5}{2 \sqrt{x}}\right) \times 5 \sqrt{x}\) \(a=\frac{5}{2} \times 5=\frac{25}{2}=12.5 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, acceleration is \(12.5 \mathrm{~m} / \mathrm{s}^{2}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141432
The acceleration (a) of an object varies as a function of its velocity \((v)\) as \(a=\lambda \sqrt{v}\) where \(\lambda\) is a constant. If at \(t=0, v=0\), then the velocity as a function of time \((t)\) is given as
1 \(\frac{3}{4} \lambda^{2} t^{3}\)
2 \(\frac{1}{4} \lambda t\)
3 \(\frac{1}{4} \lambda^{2} t^{2}\)
4 \(\frac{1}{4} \lambda^{2} t\)
Explanation:
C We know that, \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \{\mathrm{a}=\lambda \sqrt{\mathrm{v}}-\text { given }\}\) \(\mathrm{dv}=\mathrm{a} \cdot \mathrm{dt}\) Integration both side \(\int_{0}^{\mathrm{v}} \frac{\mathrm{dv}}{\sqrt{\mathrm{v}}}=\lambda \int_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}\) \({[2 \sqrt{\mathrm{v}}]_{0}^{\mathrm{v}}=\lambda[\mathrm{t}]_{0}^{\mathrm{t}}}\) \(2 \sqrt{\mathrm{v}}=\lambda[\mathrm{t}-0]\) Squaring both side \(4 v=\lambda^{2} t^{2}\) \(v=\frac{\lambda^{2} t^{2}}{4}\)
UPSEE 2020
Motion in One Dimensions
141433
Find the relation between ' \(x\) ' and ' \(y\) ', when a body starting from rest, moves with uniform acceleration and travels a distance \(x\) in the first \(\mathbf{n}\) seconds and \(\mathbf{y}\) in the next \(\mathrm{n}\) seconds:
1 \(3 x=y\)
2 \(x=3 y\)
3 \(\mathrm{x}=\mathrm{y}\)
4 \(x=2 y\)
Explanation:
A From given question. The distance covered in \(2 \mathrm{sec}, \mathrm{x}=\frac{1}{2} \mathrm{a}(2)^{2}=2 \mathrm{a}\) and the distance covered in next \(2 \mathrm{sec}\), \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} a \cdot(2)^{2}\) \(y=6 a\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 x=y \Rightarrow y=3 x\)
141429
To reach point-B from point-A, a person travels \(500 \mathrm{~m}\) North, \(400 \mathrm{~m}\) East and \(200 \mathrm{~m}\) south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person
A Distance \(=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}\) \(=(500+400+200)\) \(=1100 \mathrm{~m}\) Displacement \(=\mathrm{AD}=\sqrt{(\mathrm{AB}-\mathrm{CD})^{2}+\mathrm{BC}^{2}}\) \(=\sqrt{(500-200)^{2}+400^{2}}\) \(=500 \mathrm{~m}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(\qquad \frac{1100}{1200}=\frac{11}{12} \mathrm{~m} / \mathrm{s}\) \(\text { Average Velocity }=\frac{A D}{t}\) \(=\frac{500}{1200}\) Average velocity \(=\frac{5}{12} \mathrm{~ms}^{-1}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141430
A particle is at \(x=0\) when \(t=0\). It moves among \(x\)-axis with a velocity given by \(v=5 \sqrt{x}\). Find the acceleration of this particle
1 \(8.5 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(12.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(10.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(11.5 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
B We know that Acceleration \(=\) derivative of speed with respect to time Therefore, \(a=\frac{d v}{d t}\) \(a=\frac{d v}{d x} \times \frac{d x}{d t}\) \(a=d(5 \sqrt{x}) / d x \times \frac{d x}{d t}\) \(a=\frac{5}{2 \sqrt{x}} \times v \quad(v=5 \sqrt{x})\) \(a=\left(\frac{5}{2 \sqrt{x}}\right) \times 5 \sqrt{x}\) \(a=\frac{5}{2} \times 5=\frac{25}{2}=12.5 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, acceleration is \(12.5 \mathrm{~m} / \mathrm{s}^{2}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141432
The acceleration (a) of an object varies as a function of its velocity \((v)\) as \(a=\lambda \sqrt{v}\) where \(\lambda\) is a constant. If at \(t=0, v=0\), then the velocity as a function of time \((t)\) is given as
1 \(\frac{3}{4} \lambda^{2} t^{3}\)
2 \(\frac{1}{4} \lambda t\)
3 \(\frac{1}{4} \lambda^{2} t^{2}\)
4 \(\frac{1}{4} \lambda^{2} t\)
Explanation:
C We know that, \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \{\mathrm{a}=\lambda \sqrt{\mathrm{v}}-\text { given }\}\) \(\mathrm{dv}=\mathrm{a} \cdot \mathrm{dt}\) Integration both side \(\int_{0}^{\mathrm{v}} \frac{\mathrm{dv}}{\sqrt{\mathrm{v}}}=\lambda \int_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}\) \({[2 \sqrt{\mathrm{v}}]_{0}^{\mathrm{v}}=\lambda[\mathrm{t}]_{0}^{\mathrm{t}}}\) \(2 \sqrt{\mathrm{v}}=\lambda[\mathrm{t}-0]\) Squaring both side \(4 v=\lambda^{2} t^{2}\) \(v=\frac{\lambda^{2} t^{2}}{4}\)
UPSEE 2020
Motion in One Dimensions
141433
Find the relation between ' \(x\) ' and ' \(y\) ', when a body starting from rest, moves with uniform acceleration and travels a distance \(x\) in the first \(\mathbf{n}\) seconds and \(\mathbf{y}\) in the next \(\mathrm{n}\) seconds:
1 \(3 x=y\)
2 \(x=3 y\)
3 \(\mathrm{x}=\mathrm{y}\)
4 \(x=2 y\)
Explanation:
A From given question. The distance covered in \(2 \mathrm{sec}, \mathrm{x}=\frac{1}{2} \mathrm{a}(2)^{2}=2 \mathrm{a}\) and the distance covered in next \(2 \mathrm{sec}\), \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} a \cdot(2)^{2}\) \(y=6 a\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 x=y \Rightarrow y=3 x\)
141429
To reach point-B from point-A, a person travels \(500 \mathrm{~m}\) North, \(400 \mathrm{~m}\) East and \(200 \mathrm{~m}\) south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person
A Distance \(=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}\) \(=(500+400+200)\) \(=1100 \mathrm{~m}\) Displacement \(=\mathrm{AD}=\sqrt{(\mathrm{AB}-\mathrm{CD})^{2}+\mathrm{BC}^{2}}\) \(=\sqrt{(500-200)^{2}+400^{2}}\) \(=500 \mathrm{~m}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(\qquad \frac{1100}{1200}=\frac{11}{12} \mathrm{~m} / \mathrm{s}\) \(\text { Average Velocity }=\frac{A D}{t}\) \(=\frac{500}{1200}\) Average velocity \(=\frac{5}{12} \mathrm{~ms}^{-1}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141430
A particle is at \(x=0\) when \(t=0\). It moves among \(x\)-axis with a velocity given by \(v=5 \sqrt{x}\). Find the acceleration of this particle
1 \(8.5 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(12.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(10.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(11.5 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
B We know that Acceleration \(=\) derivative of speed with respect to time Therefore, \(a=\frac{d v}{d t}\) \(a=\frac{d v}{d x} \times \frac{d x}{d t}\) \(a=d(5 \sqrt{x}) / d x \times \frac{d x}{d t}\) \(a=\frac{5}{2 \sqrt{x}} \times v \quad(v=5 \sqrt{x})\) \(a=\left(\frac{5}{2 \sqrt{x}}\right) \times 5 \sqrt{x}\) \(a=\frac{5}{2} \times 5=\frac{25}{2}=12.5 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, acceleration is \(12.5 \mathrm{~m} / \mathrm{s}^{2}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141432
The acceleration (a) of an object varies as a function of its velocity \((v)\) as \(a=\lambda \sqrt{v}\) where \(\lambda\) is a constant. If at \(t=0, v=0\), then the velocity as a function of time \((t)\) is given as
1 \(\frac{3}{4} \lambda^{2} t^{3}\)
2 \(\frac{1}{4} \lambda t\)
3 \(\frac{1}{4} \lambda^{2} t^{2}\)
4 \(\frac{1}{4} \lambda^{2} t\)
Explanation:
C We know that, \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \{\mathrm{a}=\lambda \sqrt{\mathrm{v}}-\text { given }\}\) \(\mathrm{dv}=\mathrm{a} \cdot \mathrm{dt}\) Integration both side \(\int_{0}^{\mathrm{v}} \frac{\mathrm{dv}}{\sqrt{\mathrm{v}}}=\lambda \int_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}\) \({[2 \sqrt{\mathrm{v}}]_{0}^{\mathrm{v}}=\lambda[\mathrm{t}]_{0}^{\mathrm{t}}}\) \(2 \sqrt{\mathrm{v}}=\lambda[\mathrm{t}-0]\) Squaring both side \(4 v=\lambda^{2} t^{2}\) \(v=\frac{\lambda^{2} t^{2}}{4}\)
UPSEE 2020
Motion in One Dimensions
141433
Find the relation between ' \(x\) ' and ' \(y\) ', when a body starting from rest, moves with uniform acceleration and travels a distance \(x\) in the first \(\mathbf{n}\) seconds and \(\mathbf{y}\) in the next \(\mathrm{n}\) seconds:
1 \(3 x=y\)
2 \(x=3 y\)
3 \(\mathrm{x}=\mathrm{y}\)
4 \(x=2 y\)
Explanation:
A From given question. The distance covered in \(2 \mathrm{sec}, \mathrm{x}=\frac{1}{2} \mathrm{a}(2)^{2}=2 \mathrm{a}\) and the distance covered in next \(2 \mathrm{sec}\), \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} a \cdot(2)^{2}\) \(y=6 a\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 x=y \Rightarrow y=3 x\)
141429
To reach point-B from point-A, a person travels \(500 \mathrm{~m}\) North, \(400 \mathrm{~m}\) East and \(200 \mathrm{~m}\) south. If the person takes 1200 seconds to reach point-B from point-A then find the average velocity of the person
A Distance \(=\mathrm{AB}+\mathrm{BC}+\mathrm{CD}\) \(=(500+400+200)\) \(=1100 \mathrm{~m}\) Displacement \(=\mathrm{AD}=\sqrt{(\mathrm{AB}-\mathrm{CD})^{2}+\mathrm{BC}^{2}}\) \(=\sqrt{(500-200)^{2}+400^{2}}\) \(=500 \mathrm{~m}\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(\qquad \frac{1100}{1200}=\frac{11}{12} \mathrm{~m} / \mathrm{s}\) \(\text { Average Velocity }=\frac{A D}{t}\) \(=\frac{500}{1200}\) Average velocity \(=\frac{5}{12} \mathrm{~ms}^{-1}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141430
A particle is at \(x=0\) when \(t=0\). It moves among \(x\)-axis with a velocity given by \(v=5 \sqrt{x}\). Find the acceleration of this particle
1 \(8.5 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(12.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
3 \(10.5 \mathrm{~m} \cdot \mathrm{s}^{-2}\)
4 \(11.5 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
B We know that Acceleration \(=\) derivative of speed with respect to time Therefore, \(a=\frac{d v}{d t}\) \(a=\frac{d v}{d x} \times \frac{d x}{d t}\) \(a=d(5 \sqrt{x}) / d x \times \frac{d x}{d t}\) \(a=\frac{5}{2 \sqrt{x}} \times v \quad(v=5 \sqrt{x})\) \(a=\left(\frac{5}{2 \sqrt{x}}\right) \times 5 \sqrt{x}\) \(a=\frac{5}{2} \times 5=\frac{25}{2}=12.5 \mathrm{~m} / \mathrm{s}^{2}\) Therefore, acceleration is \(12.5 \mathrm{~m} / \mathrm{s}^{2}\)
AP EAMCET-05.10.2021
Motion in One Dimensions
141432
The acceleration (a) of an object varies as a function of its velocity \((v)\) as \(a=\lambda \sqrt{v}\) where \(\lambda\) is a constant. If at \(t=0, v=0\), then the velocity as a function of time \((t)\) is given as
1 \(\frac{3}{4} \lambda^{2} t^{3}\)
2 \(\frac{1}{4} \lambda t\)
3 \(\frac{1}{4} \lambda^{2} t^{2}\)
4 \(\frac{1}{4} \lambda^{2} t\)
Explanation:
C We know that, \(\mathrm{a}=\frac{\mathrm{dv}}{\mathrm{dt}} \{\mathrm{a}=\lambda \sqrt{\mathrm{v}}-\text { given }\}\) \(\mathrm{dv}=\mathrm{a} \cdot \mathrm{dt}\) Integration both side \(\int_{0}^{\mathrm{v}} \frac{\mathrm{dv}}{\sqrt{\mathrm{v}}}=\lambda \int_{\mathrm{t}=0}^{\mathrm{t}} \mathrm{dt}\) \({[2 \sqrt{\mathrm{v}}]_{0}^{\mathrm{v}}=\lambda[\mathrm{t}]_{0}^{\mathrm{t}}}\) \(2 \sqrt{\mathrm{v}}=\lambda[\mathrm{t}-0]\) Squaring both side \(4 v=\lambda^{2} t^{2}\) \(v=\frac{\lambda^{2} t^{2}}{4}\)
UPSEE 2020
Motion in One Dimensions
141433
Find the relation between ' \(x\) ' and ' \(y\) ', when a body starting from rest, moves with uniform acceleration and travels a distance \(x\) in the first \(\mathbf{n}\) seconds and \(\mathbf{y}\) in the next \(\mathrm{n}\) seconds:
1 \(3 x=y\)
2 \(x=3 y\)
3 \(\mathrm{x}=\mathrm{y}\)
4 \(x=2 y\)
Explanation:
A From given question. The distance covered in \(2 \mathrm{sec}, \mathrm{x}=\frac{1}{2} \mathrm{a}(2)^{2}=2 \mathrm{a}\) and the distance covered in next \(2 \mathrm{sec}\), \(y=\frac{1}{2} \cdot a(4)^{2}-\frac{1}{2} a \cdot(2)^{2}\) \(y=6 a\) So, \(\frac{x}{y}=\frac{2 a}{6 a}=\frac{1}{3}\) \(3 x=y \Rightarrow y=3 x\)
