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 m/s²) 

(1) 24.5 m/s

(2) 49.0 m/s

(3) 73.5 m/s

(4) 98.0 m/s

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?

(1) 293 m

(2) 111 m

(3) 91 m

(4) 182 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 speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is:  

1. 4.0 m/s

2. 5.0 m/s

3. 5.5 m/s

4. 4.8 m/s

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 m/s

2. 55 m/s

3. 550 m/s

4. 660 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 t₁ and t₂ 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. $t=t_1^2{t}_2^2$