If a freely falling body travels in the last second a distance equal to the distance travelled by it in the first three seconds, the time of the travel is:
1. \(6\) sec
2. \(5\) sec
3. \(4\) sec
4. \(3\) sec

The effective acceleration of a body, when thrown upwards with acceleration a will be: 

1. $\sqrt{a-g^2}$

2. $\sqrt{a^2+g^2}$

3. $(a-g)$

4. $(a+g)$

A body is thrown vertically upwards with velocity \(u.\) The distance travelled by it in the fifth and the sixth seconds are equal. The velocity \(u\) is given by (\(g = 9.8~\text{m/s}^2\)) 

1. \(24.5~\text{m/s}\)
2. \(49.0~\text{m/s}\)
3. \(73.5~\text{m/s}\)
4. \(98.0~\text{m/s}\)

A parachutist after bailing out falls \(50~\text{m}\) without friction. When parachute opens, it decelerates at \(2~\text{m/s}^2\). He reaches the ground with a speed of \(3~\text{m/s}\). At what height, did he bail out ?
1. \(293~\text{m}\)
2. \(111~\text{m}\)
3. \(91~\text{m}\)
4. \(182~\text{m}\)

When a ball is thrown up vertically with velocity v₀, it reaches a maximum height of 'h'. If one wishes to triple the maximum height then the ball should be thrown with velocity

1. $\sqrt{3}{v}_o$

2. 3v₀

3. 9v₀

4. (3/2)v₀

A particle moving in a straight line covers half the distance with a speed of \(3~\text{m/s}\). The other half of the distance is covered in two equal time intervals with speeds of \(4.5~\text{m/s}\) and \(7.5~\text{m/s}\) respectively. The average speed of the particle during this motion is:

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity of v₀. The distance travelled by the particle in time t will be:

(1) $v_{0}t+\frac{1}{3}b{t}^2$

(2) $v_{0}t+\frac{1}{3}b{t}^3$

(3) $v_{0}t+\frac{1}{6}b{t}^3$

(4) $v_{0}t+\frac{1}{2}b{t}^2$

A particle starts from rest. Its acceleration \((a)\) versus time \((t)\) is as shown in the figure. The maximum speed of the particle will be: 

             
1. \(110~\text{m/s}\)
2. \(55~\text{m/s}\)
3. \(550~\text{m/s}\)
4. \(660~\text{m/s}\)

A car accelerates from rest at a constant rate α for some time, after which it decelerates at a constant rate β and comes to rest. If the total time elapsed is t, then the maximum velocity acquired by the car is

1.  $\left(\frac{{\displaystyle {\alpha }^2+{\beta }^2}}{{\displaystyle \alpha \beta }}\right)t$
2.  $\left(\frac{{\displaystyle {\alpha }^2-{\beta }^2}}{{\displaystyle \alpha \beta }}\right)\hspace{0.17em}t$
3.  $\frac{(\alpha +\beta )t}{\alpha \beta }$
4.  $\frac{\alpha \beta t}{\alpha +\beta }$

A stone dropped from a building of height \(h\) and reaches the earth after \(t\) seconds. From the same building, if two stones are thrown (one upwards and other downwards) with the same velocity \(u\) and they reach the earth surface after \(t_1\) and \(t_2\) seconds respectively, then: 

1. $t=t_1-t_2$

2. $t=\frac{t_1+t_2}{2}$

3. $t=\sqrt{t_1{t}_2}$

4. $t=t_1^2{t}_2^2$