The displacement of a particle moving in a straight line is given by \(𝑠 = 2 𝑡^2 + 2 𝑡 + 4\) where \(s\) is in meters and \(t\) in seconds. The acceleration of the particle is:

1. \(2\ \text{m/s}^2\)
2. \(4\ \text{m/s}^2\)
3. \(6\ \text{m/s}^2\)
4. \(8\ \text{m/s}^2\)

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 particle starts from rest, accelerates at \(2\ \text{m/s}^2\) for \(10\ \text{s}\) and then goes for constant speed for \(30\ \text{s}\) and then decelerates at \(4\ \text{m/s}^2\) till it stops. What is the distance travelled by it?

1. \(750\ \text{m}\)
2. \(800\ \text{m}\)
3. \(700\ \text{m}\)
4. \(850\ \text{m}\)

A car, moving with a speed of \(50\ \text{km/h}\), can be stopped by the brakes after at least \(6\ \text{m}\). If the same car is moving at a speed of \(100\ \text{km/h}\), the minimum stopping distance is:

1. \(6\ \text{m}\)
2. \(12\ \text{m}\)
3. \(18\ \text{m}\)
4. \(24\ \text{m}\)

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

A body \(A\) moves with a uniform acceleration \(a\) and zero initial velocity. Another body \(B\), starts from the same point and moves in the same direction with a constant velocity \(v\). The two bodies meet after a time \(t\). The value of \(t\) is: 

1. \(\dfrac{2v}{a}\)
 
2. \(\dfrac{v}{a}\)

3. \(\dfrac{v}{2a}\)

4. \(\sqrt{\dfrac{v}{2a}}\)

A particle moves along \(x\)-axis in such a way that its coordinate \(x\) varies with time \(t\) according to the equation \(𝑥 = ( 2 − 5 𝑡 + 6 𝑡^ 2 )  \) m. The initial velocity of the particle is:

1. \(–5\ \text{m/s}\)

2. \(6\ \text{m/s}\)

3. \(–3\ \text{m/s}\)

4. \(3\ \text{m/s}\)

A car starts from rest and moves with uniform acceleration \(a\) on a straight road from time \(t = 0\) to \(t = T\). After that, a constant deceleration brings it to rest. In this process, the average speed of the car is: 

1. \(\dfrac{aT}{4}\)

2. \(\dfrac{3aT}{2}\)

3. \(\dfrac{aT}{2}\)

4. \(aT\)

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. \(275\ \text{m}\)

If the velocity of a particle is given by \(v = (180-16x)^{1/2}~\text{m/s} \), then its acceleration will be: 
1. zero
2. \(8\text{ m/s}^2\)
3. \(-8\text{ m/s}^2\)
4. \(4\text{ m/s}^2\)