The velocity of a bullet is reduced from \(200 \ \text{m/s}\) to \(100 \ \text{m/s}\) while travelling through a wooden block of thickness \(10\ \text{cm}\). The retardation, assuming it to be uniform, will be: 

1. \(10×10^4\ \text{m/s}^2\)

2. \(12×10^4\ \text{m/s}^2\)$\mathrm{}$

3. \(13.5×10^4\ \text{m/s}^2\)$\mathrm{}$

4. \(15×10^4\ \text{m/s}^2\)$\mathrm{}$

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

An object accelerates from rest to a velocity of \(27.5\ \text{m/s}\) in \(10\ \text{s}\). Then find the distance covered by the object in the next \(10\ \text{s}\):

1. \(550\ \text{m}\)
2. \(137.5\ \text{m}\)
3. \(412.5\ \text{m}\)
4. \(275\ \text{m}\)

The speeds of two identical cars are \(u\) and \(4u\) at a specific instant. The ratio of the respective distances in which the two cars are stopped from that instant is:

1. \(1 : 1\)
2. \(1 : 4\)
3. \(1 : 8\)
4. \(1 : 16\)

A car, starting from rest, accelerates at the rate \(f\) through a distance \(S\), then continues at a constant speed for time \(t\) and then decelerates at the rate \(\frac f2\) to come to rest. If the total distance traversed is \(15\ \text{S}\), then:

1. \(S = \frac{1}{2}ft^2\)
2. \(S = \frac{1}{4}ft^2\)
3. \(S = \frac{1}{72}ft^2\)
4. \(S = \frac{1}{6}ft^2\)

A man is \(45\ \text{m}\) behind the bus when the bus starts accelerating from rest with an acceleration of \(2.5\ \text{m/s}^2\). With what minimum velocity should the man start running to catch the bus?

1.  \(12\ \text{m/s}\)
2.  \(14\ \text{m/s}\)
3.  \(15\ \text{m/s}\)
4.  \(16\ \text{m/s}\)

A \(120\ \text{m}\) long train is moving in a direction with speed \(20\ \text{m/s}\). A train \(B\), moving with \(30\ \text{m/s}\) in the opposite direction and \(130\ \text{m}\) long, crosses the first train in a time:

1.  \(4\ \text{s}\)
2.  \(36\ \text{s}\)
3.  \(38\ \text{s}\)
4.  \(5\ \text{s}\)

A \(210\) meter long train is moving due north at a speed of \(25\ \text{m/s}\). A small bird is flying due South a little above the train with a speed of \(5\ \text{m/s}\). The time taken by the bird to cross the train is:

1. \(6\ \text{s}\)
2. \(7\ \text{s}\)
3. \(9\ \text{s}\)
4. \(10\ \text{s}\)

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 man in a balloon rising vertically with an acceleration of $4.9m/{\sec }^2$ releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is $(g=9.8m/{\sec }^2)$  

1. 14.7 m

2. 19.6 m

3. 9.8 m

4. 24.5 m