141423
A train travels from city-A to city-B with a constant speed of \(18 \mathrm{~m} . \mathrm{s}^{-1}\) and returns back to city-A with a constant speed of \(36 \mathrm{~m}^{-1}\). Find its average speed during the journey.
C Let the distance between the two cities A and \(\mathrm{B}\) is \(\mathrm{x}\) meter. Time taken by the train to travel from A to B \(\frac{\mathrm{x}}{18}=\mathrm{t}_1\left(\begin{array}{l}\because \text { Time }=\frac{\text { Distance }}{\text { Speed }} \\ \text { A to Bspeed }=18 \mathrm{~ms}^{-1}\end{array}\right)\) Time taken to come back from B to A. \(\frac{\mathrm{X}}{36}=\mathrm{t}_{2}\left(\text { B to A speed }=36 \mathrm{~ms}^{-1}\right)\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{x+x}{t_{1}+t_{2}}\) \(=\frac{2 x}{\frac{x}{18}+\frac{x}{36}}=\frac{2 x \times 36}{2 x+x}\) \(=\frac{2 x \times 36}{3 x}=\frac{72 x}{3 x} \mathrm{~m} / \mathrm{sec}\) Average speed \(=\frac{72}{3} \mathrm{~ms}^{-1}\)
AP EAMCET-24.08.2021
Motion in One Dimensions
141424
An object travelling at a speed of \(36 \mathrm{kmph}\) comes to rest in a distance of \(200 \mathrm{~m}\) after the brakes were applied. The retardation produced by the brakes is
1 \(0.25 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(0.20 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(0.15 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(0.10 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
A Given that, \(\mathrm{u}=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~ms}^{-1}\) \(\mathrm{~s}=200 \mathrm{~m}\) \(\mathrm{v}=0\) \(\mathrm{a}=?\) By third law of Motion- \(v^{2}-u^{2}=2 a s\) \(0^{2}-(10)^{2}=2 \times a \times 200\) \(-100=400 a\) \(a=\frac{-100}{400}=-0.25\) Retardation is \(=0.25 \mathrm{~ms}^{-2}\)
AP EAMCET-19.08.2021
Motion in One Dimensions
141428
The velocity of a particle is given by \(v=2 t^{2}-8 t\) \(+15 \mathrm{~ms}^{-1}\). Find its instantaneous acceleration at \(\mathbf{t}=\mathbf{5 s}\).
141426
Assertion (A) : An object can possess acceleration even at a time when it has a uniform speed. Reason (R) : It is possible when the direction of motion keeps changing.
1 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is a correct explanation for \(\mathrm{A}\)
2 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not a correct explanation for \(\mathrm{A}\)
3 \(\mathrm{A}\) is true. \(\mathrm{R}\) is false
4 \(\mathrm{A}\) is false. \(\mathrm{R}\) is true
Explanation:
A Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is correct explanation for \(\mathrm{A}\) We know that, An object can possess acceleration even at a time when it has uniform speed because, it is possible when the direction of motion keeps changing. e.g. Uniform circular motion
141423
A train travels from city-A to city-B with a constant speed of \(18 \mathrm{~m} . \mathrm{s}^{-1}\) and returns back to city-A with a constant speed of \(36 \mathrm{~m}^{-1}\). Find its average speed during the journey.
C Let the distance between the two cities A and \(\mathrm{B}\) is \(\mathrm{x}\) meter. Time taken by the train to travel from A to B \(\frac{\mathrm{x}}{18}=\mathrm{t}_1\left(\begin{array}{l}\because \text { Time }=\frac{\text { Distance }}{\text { Speed }} \\ \text { A to Bspeed }=18 \mathrm{~ms}^{-1}\end{array}\right)\) Time taken to come back from B to A. \(\frac{\mathrm{X}}{36}=\mathrm{t}_{2}\left(\text { B to A speed }=36 \mathrm{~ms}^{-1}\right)\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{x+x}{t_{1}+t_{2}}\) \(=\frac{2 x}{\frac{x}{18}+\frac{x}{36}}=\frac{2 x \times 36}{2 x+x}\) \(=\frac{2 x \times 36}{3 x}=\frac{72 x}{3 x} \mathrm{~m} / \mathrm{sec}\) Average speed \(=\frac{72}{3} \mathrm{~ms}^{-1}\)
AP EAMCET-24.08.2021
Motion in One Dimensions
141424
An object travelling at a speed of \(36 \mathrm{kmph}\) comes to rest in a distance of \(200 \mathrm{~m}\) after the brakes were applied. The retardation produced by the brakes is
1 \(0.25 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(0.20 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(0.15 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(0.10 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
A Given that, \(\mathrm{u}=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~ms}^{-1}\) \(\mathrm{~s}=200 \mathrm{~m}\) \(\mathrm{v}=0\) \(\mathrm{a}=?\) By third law of Motion- \(v^{2}-u^{2}=2 a s\) \(0^{2}-(10)^{2}=2 \times a \times 200\) \(-100=400 a\) \(a=\frac{-100}{400}=-0.25\) Retardation is \(=0.25 \mathrm{~ms}^{-2}\)
AP EAMCET-19.08.2021
Motion in One Dimensions
141428
The velocity of a particle is given by \(v=2 t^{2}-8 t\) \(+15 \mathrm{~ms}^{-1}\). Find its instantaneous acceleration at \(\mathbf{t}=\mathbf{5 s}\).
141426
Assertion (A) : An object can possess acceleration even at a time when it has a uniform speed. Reason (R) : It is possible when the direction of motion keeps changing.
1 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is a correct explanation for \(\mathrm{A}\)
2 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not a correct explanation for \(\mathrm{A}\)
3 \(\mathrm{A}\) is true. \(\mathrm{R}\) is false
4 \(\mathrm{A}\) is false. \(\mathrm{R}\) is true
Explanation:
A Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is correct explanation for \(\mathrm{A}\) We know that, An object can possess acceleration even at a time when it has uniform speed because, it is possible when the direction of motion keeps changing. e.g. Uniform circular motion
141423
A train travels from city-A to city-B with a constant speed of \(18 \mathrm{~m} . \mathrm{s}^{-1}\) and returns back to city-A with a constant speed of \(36 \mathrm{~m}^{-1}\). Find its average speed during the journey.
C Let the distance between the two cities A and \(\mathrm{B}\) is \(\mathrm{x}\) meter. Time taken by the train to travel from A to B \(\frac{\mathrm{x}}{18}=\mathrm{t}_1\left(\begin{array}{l}\because \text { Time }=\frac{\text { Distance }}{\text { Speed }} \\ \text { A to Bspeed }=18 \mathrm{~ms}^{-1}\end{array}\right)\) Time taken to come back from B to A. \(\frac{\mathrm{X}}{36}=\mathrm{t}_{2}\left(\text { B to A speed }=36 \mathrm{~ms}^{-1}\right)\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{x+x}{t_{1}+t_{2}}\) \(=\frac{2 x}{\frac{x}{18}+\frac{x}{36}}=\frac{2 x \times 36}{2 x+x}\) \(=\frac{2 x \times 36}{3 x}=\frac{72 x}{3 x} \mathrm{~m} / \mathrm{sec}\) Average speed \(=\frac{72}{3} \mathrm{~ms}^{-1}\)
AP EAMCET-24.08.2021
Motion in One Dimensions
141424
An object travelling at a speed of \(36 \mathrm{kmph}\) comes to rest in a distance of \(200 \mathrm{~m}\) after the brakes were applied. The retardation produced by the brakes is
1 \(0.25 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(0.20 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(0.15 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(0.10 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
A Given that, \(\mathrm{u}=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~ms}^{-1}\) \(\mathrm{~s}=200 \mathrm{~m}\) \(\mathrm{v}=0\) \(\mathrm{a}=?\) By third law of Motion- \(v^{2}-u^{2}=2 a s\) \(0^{2}-(10)^{2}=2 \times a \times 200\) \(-100=400 a\) \(a=\frac{-100}{400}=-0.25\) Retardation is \(=0.25 \mathrm{~ms}^{-2}\)
AP EAMCET-19.08.2021
Motion in One Dimensions
141428
The velocity of a particle is given by \(v=2 t^{2}-8 t\) \(+15 \mathrm{~ms}^{-1}\). Find its instantaneous acceleration at \(\mathbf{t}=\mathbf{5 s}\).
141426
Assertion (A) : An object can possess acceleration even at a time when it has a uniform speed. Reason (R) : It is possible when the direction of motion keeps changing.
1 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is a correct explanation for \(\mathrm{A}\)
2 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not a correct explanation for \(\mathrm{A}\)
3 \(\mathrm{A}\) is true. \(\mathrm{R}\) is false
4 \(\mathrm{A}\) is false. \(\mathrm{R}\) is true
Explanation:
A Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is correct explanation for \(\mathrm{A}\) We know that, An object can possess acceleration even at a time when it has uniform speed because, it is possible when the direction of motion keeps changing. e.g. Uniform circular motion
141423
A train travels from city-A to city-B with a constant speed of \(18 \mathrm{~m} . \mathrm{s}^{-1}\) and returns back to city-A with a constant speed of \(36 \mathrm{~m}^{-1}\). Find its average speed during the journey.
C Let the distance between the two cities A and \(\mathrm{B}\) is \(\mathrm{x}\) meter. Time taken by the train to travel from A to B \(\frac{\mathrm{x}}{18}=\mathrm{t}_1\left(\begin{array}{l}\because \text { Time }=\frac{\text { Distance }}{\text { Speed }} \\ \text { A to Bspeed }=18 \mathrm{~ms}^{-1}\end{array}\right)\) Time taken to come back from B to A. \(\frac{\mathrm{X}}{36}=\mathrm{t}_{2}\left(\text { B to A speed }=36 \mathrm{~ms}^{-1}\right)\) Average speed \(=\frac{\text { Total distance }}{\text { Total time }}\) \(=\frac{x+x}{t_{1}+t_{2}}\) \(=\frac{2 x}{\frac{x}{18}+\frac{x}{36}}=\frac{2 x \times 36}{2 x+x}\) \(=\frac{2 x \times 36}{3 x}=\frac{72 x}{3 x} \mathrm{~m} / \mathrm{sec}\) Average speed \(=\frac{72}{3} \mathrm{~ms}^{-1}\)
AP EAMCET-24.08.2021
Motion in One Dimensions
141424
An object travelling at a speed of \(36 \mathrm{kmph}\) comes to rest in a distance of \(200 \mathrm{~m}\) after the brakes were applied. The retardation produced by the brakes is
1 \(0.25 \mathrm{~m} . \mathrm{s}^{-2}\)
2 \(0.20 \mathrm{~m} . \mathrm{s}^{-2}\)
3 \(0.15 \mathrm{~m} . \mathrm{s}^{-2}\)
4 \(0.10 \mathrm{~m} . \mathrm{s}^{-2}\)
Explanation:
A Given that, \(\mathrm{u}=36 \mathrm{~km} / \mathrm{h}=36 \times \frac{5}{18}=10 \mathrm{~ms}^{-1}\) \(\mathrm{~s}=200 \mathrm{~m}\) \(\mathrm{v}=0\) \(\mathrm{a}=?\) By third law of Motion- \(v^{2}-u^{2}=2 a s\) \(0^{2}-(10)^{2}=2 \times a \times 200\) \(-100=400 a\) \(a=\frac{-100}{400}=-0.25\) Retardation is \(=0.25 \mathrm{~ms}^{-2}\)
AP EAMCET-19.08.2021
Motion in One Dimensions
141428
The velocity of a particle is given by \(v=2 t^{2}-8 t\) \(+15 \mathrm{~ms}^{-1}\). Find its instantaneous acceleration at \(\mathbf{t}=\mathbf{5 s}\).
141426
Assertion (A) : An object can possess acceleration even at a time when it has a uniform speed. Reason (R) : It is possible when the direction of motion keeps changing.
1 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is a correct explanation for \(\mathrm{A}\)
2 Both \(\mathrm{A}\) and \(\mathrm{R}\) are true but \(\mathrm{R}\) is not a correct explanation for \(\mathrm{A}\)
3 \(\mathrm{A}\) is true. \(\mathrm{R}\) is false
4 \(\mathrm{A}\) is false. \(\mathrm{R}\) is true
Explanation:
A Both \(\mathrm{A}\) and \(\mathrm{R}\) are true and \(\mathrm{R}\) is correct explanation for \(\mathrm{A}\) We know that, An object can possess acceleration even at a time when it has uniform speed because, it is possible when the direction of motion keeps changing. e.g. Uniform circular motion
