A point moves in a straight line under the retardation a. If the initial velocity is \(\mathrm{u},\) the distance covered in \(\mathrm{t}\) seconds is:
1.
2.
3.
4.
The relation between time and distance is given by , where α and β are constants. The retardation, as calculated based on this equation, will be (assume v to be velocity) :
1.
2.
3.
4.
The displacement of a particle is given by . The initial velocity and acceleration are, respectively:
1. | \(\mathrm{b}, ~\mathrm{-4d}\) | 2. | \(\mathrm{-b},~ \mathrm{2c}\) |
3. | \(\mathrm{b}, ~\mathrm{2c}\) | 4. | \(\mathrm{2c}, ~\mathrm{-2d}\) |
The acceleration ‘a’ in m/s2 of a particle is given by where t is the time. If the particle starts out with a velocity, u = 2 m/s at t = 0, then the velocity at the end of 2 seconds will be:
1. 12 m/s
2. 18 m/s
3. 27 m/s
4. 36 m/s
A particle moves along a straight line such that its displacement at any time t is given by metres. The velocity when the acceleration is zero is:
1. | 4 ms-1 | 2. | −12 ms−1 |
3. | 42 ms−1 | 4. | −9 ms−1 |
The position \(x\) of a particle varies with time \(t\) as \(x=at^2-bt^3\). The acceleration of the particle will be zero at time \(t\) equal to:
1. \(\frac{a}{b}\)
2. \(\frac{2a}{3b}\)
3. \(\frac{a}{3b}\)
4. zero
A student is standing at a distance of 50 metres from the bus. As soon as the bus begins its motion with an acceleration of 1 ms–2, the student starts running towards the bus with a uniform velocity u. Assuming the motion to be along a straight road, the minimum value of u, so that the student is able to catch the bus is:
1. 5 ms–1
2. 8 ms–1
3. 10 ms–1
4. 12 ms–1
If the velocity of a particle is given by m/s, then its acceleration will be:
1. Zero
2. 8 m/s2
3. – 8 m/s2
4. 4 m/s2
A body is thrown vertically upwards. If the air resistance is to be taken into account, then the time during which the body rises is:
1. | Equal to the time of fall |
2. | Less than the time of fall |
3. | Greater than the time of fall |
4. | Twice the time of fall |
A body starts to fall freely under gravity. The distances covered by it in the first, second and third second will be in the ratio:
1. \(1:3:5\)
2. \(1:2:3\)
3. \(1:4:9\)
4. \(1:5:6\)