Two trains, each 50 m long, are travelling in the opposite direction with velocities 10 m/s and 15 m/s. The time of crossing is:

1. 10 sec

2. 4 sec

3. $2\sqrt{3}$ $\sec$

4. $4\sqrt{3}$ $\sec$

The distance between two particles is decreasing at the rate of 6 m/sec when they are moving in the opposite directions. If these particles travel with the same initial speeds and in the same direction, then the separation increases at the rate of 4 m/sec. It can be concluded that particles' speeds could be

1. 5 m/sec, 1 m/sec

2. 4 m/sec, 1 m/sec

3. 4 m/sec, 2 m/sec

4. 5 m/sec, 2 m/sec

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: 

A body starts to fall freely under gravity. The distances covered by it in 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

A body is thrown vertically up from the ground. It reaches a maximum height of 100 m in 5 seconds. After what time will it reach the ground from the position of maximum height?
1. 1.2 sec
2. 5 sec
3. 10 sec
4. 25 sec

If a body is thrown up with the velocity of 15 m/s, then the maximum height attained by the body is: (assume g = 10 m/s²) 

1. 11.25 m

2. 16.2 m

3. 24.5 m

4. 7.62 m

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

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:
1. \(4.0~\text{m/s}\)
2. \(5.0~\text{m/s}\)
3. \(5.5~\text{m/s}\)
4. \(4.8~\text{m/s}\)

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