A car moves from \(\mathrm{X}\) to \(\mathrm{Y}\) with a uniform speed \(\mathrm{v_u}\) and returns to \(\mathrm{X}\) with a uniform speed \(\mathrm{v_d}.\) The average speed for this round trip is:

1. $\frac{2{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

2. $\sqrt{{\mathrm{v}}_{\mathrm{u}}{\mathrm{v}}_{\mathrm{d}}}$

3. $\frac{{\mathrm{v}}_{\mathrm{d}}{\mathrm{v}}_{\mathrm{u}}}{{\mathrm{v}}_{\mathrm{d}}+{\mathrm{v}}_{\mathrm{u}}}$

4. $\frac{{\mathrm{v}}_{\mathrm{u}}+{\mathrm{v}}_{\mathrm{d}}}{2}$

A particle moves along a straight line OX. At a time t (in seconds), the displacement x (in metres) of the particle from O is given by x= 40 + 12t – t³. How long would the particle travel before coming to rest?

A ball is dropped vertically from a height h above the ground. It hits the ground and bounces up vertically to a height of h/2. Neglecting subsequent motion and air resistance, its velocity v varies with the height h as:
[Take vertically upwards direction as positive.]

1.		2.
3.		4.

A car A is traveling on a straight level road at a uniform speed of 60 km/h. It is followed by another car B which is moving at a speed of 70 km/h. When the distance between them is 2.5 km, car B is given a deceleration of 20 km/h². After how much time will car B catch up with car A?

1. 1 hr

2. 1/2 hr

3. 1/4 hr

4. 1/8 hr

The graph of displacement time is given below.

Its corresponding velocity-time graph will be:

1.		2.
3.		4.

A particle moves in a straight line with a constant acceleration. It changes its velocity from \(10\) ms^-1 to \(20\) ms^-1 while covering a distance of \(135\) m in \(t\) seconds. The value of \(t\) is:

A ball is dropped from a high-rise platform at t = 0 starting from rest. After 6 seconds, another ball is thrown downwards from the same platform with speed v. The two balls meet after 18 seconds. What is the value of v?

A stone falls freely under gravity. It covers distances \(h_1,~h_2\) and \(h_3\) in the first \(5\) seconds, the next \(5\) seconds and the next \(5\) seconds respectively. The relation between \(h_1,~h_2\) and \(h_3\) is:

A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to $\mathrm{v}\left(\mathrm{x}\right)$ $={\mathrm{\beta x}}^{-2\mathrm{n}}$ where $\mathrm{\beta }$ and \(\mathrm{n}\) are constants and \(\mathrm{x}\) is the position of the particle. The acceleration of the particle as a function of \(\mathrm{x}\) is given by:
1. $-2{\mathrm{n\beta }}^2{\mathrm{x}}^{-2\mathrm{n}-1}$
2. $-2{\mathrm{n\beta }}^2{\mathrm{x}}^{-4\mathrm{n}-1}$
3. $-2{\mathrm{\beta }}^2{\mathrm{x}}^{-2\mathrm{n}+1}$
4. $-2{\mathrm{n\beta }}^2{\mathrm{x}}^{-4\mathrm{n}+1}$

If the velocity of a particle is \(v=At+Bt^{2},\) where \(A\) and \(B\) are constants, then the distance travelled by it between \(1~\text{s}\) and \(2~\text{s}\) is: