- 1(current)
Q. No.The position of a particle as a function of time \(t\), is given by;
\(x(t)=a t+b t^2-c t^3\)
where \(a\), \(b\) and \(c\) are constants. When the particle attains zero acceleration, then its velocity will be:
1. \( a+\frac{b^2}{4 c} \)
2. \( a+\frac{b^2}{c} \)
3. \( a+\frac{b^2}{3 c} \)
4. \( a+\frac{b^2}{2 c}\)
Subtopic: Acceleration |
84%
From NCERT
JEE
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A particle is moving in a straight line such that its velocity is increasing at \(5\) ms-1 per meter. The acceleration of the particle at a point where its velocity is \(20\) ms-1, is:
1. \(100\) ms-2
2. \(200\) ms-2
3. \(300\) ms-2
4. \(400\) ms-2
1. \(100\) ms-2
2. \(200\) ms-2
3. \(300\) ms-2
4. \(400\) ms-2
Subtopic: Acceleration |
76%
From NCERT
JEE
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A train (moving with initial speed = \(20\) m/s) applies brakes to stop at the incoming station which is \(500\) m ahead. If brakes are applied after moving \(250\) m, then how much beyond the station train would stop?
1. \(125\) m
2. \(500\) m
3. \(250\) m
4. \(400\) m
1. \(125\) m
2. \(500\) m
3. \(250\) m
4. \(400\) m
Subtopic: Acceleration |
From NCERT
JEE
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Hints
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Given below are two statements:
| Statement I: | We can get displacement from the acceleration-time graph. |
| Statement II: | We can get acceleration from the velocity-time graph. |
| 1. | Both Statement I and Statement II are correct. |
| 2. | Both Statement I and Statement II are incorrect. |
| 3. | Statement I is correct and Statement II is incorrect. |
| 4. | Statement I is incorrect and Statement II is correct. |
Subtopic: Acceleration |
70%
From NCERT
JEE
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- 1(current)
Q. No.
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