## The velocity of a particle moving in the positive direction of the X – axis varies as v = K √s where K is a positive constant.

Q: The velocity of a particle moving in the positive direction of the X – axis varies as v = K √s where K is a positive constant. v varies with time as

Sol: $\large v = K \sqrt{s}$

$\large \frac{ds}{dt} = K \sqrt{s}$

$\large \int_{0}^{s}\frac{ds}{\sqrt{s}} = K \int_{0}^{t} dt$

$\large 2 \sqrt{s} = K t$

$\large s = \frac{1}{4} K^2 t^2$

$\large v = \frac{ds}{dt} = \frac{1}{2} K^2 t$

$\large v \propto t$

The v – t graph is a straight line passing through the origin.

## An object moving with a speed of 6.25 m/s, is decelerated at a rage given by dv/dt =-2.5√v , where v is instantaneous speed.

Q: An object moving with a speed of 6.25 m/s, is decelerated at a rage given by dv/dt =-2.5√v , where v is instantaneous speed. The time taken by the object, to come to rest, would be

Sol:$\large \frac{dv}{dt} = -2.5 \sqrt{v}$

$\large \frac{dv}{\sqrt{v} } = -2.5 dt$

$\large \int_{6.25}^{0}\frac{dv}{\sqrt{v}} = -2.5 \int_{0}^{t} dt$

$\large 2[v^{1/2}]_{6.25}^{0} = -2.5 t$

t = 2 sec

## The acceleration (a) of a particle moving in a straight line varies with its displacement (S) as a = 2S…

Q: The acceleration (a) of a particle moving in a straight line varies with its displacement (S) as a = 2 s . The velocity of the particle is zero at zero displacement. Find the corresponding velocity – displacement equation.

Sol: a = 2 s

$\large v\frac{dv}{ds} = 2 s$

$\large \int_{0}^{v}v dv = \int_{0}^{s} 2 s ds$

$\large [\frac{v^2}{2}]_{0}^{v} = 2 [\frac{s^2}{2}]_{0}^{s}$

$\large \frac{v^2}{2} = s^2$

## A particle located at x = 0, at time t = 0, starts moving along the positive x – direction with a velocity v that varies…

Q: A particle located at x = 0, at time t = 0, starts moving along the positive x – direction with a velocity v that varies as v = α√x. The displacement of particle varies with time as

Sol: $\large v = \alpha \sqrt{x}$

$\large \frac{dx}{dt} = \alpha \sqrt{x}$

$\large \frac{dx}{\sqrt{x}} = \alpha dt$

On integrating ,

$\large \int_{0}^{x} \frac{dx}{\sqrt{x}} = \int_{0}^{t} \alpha dt$

$\large [\frac{x^{1/2}}{1/2}]_{0}^{x} = \alpha t$

$\large x^{1/2} = \frac{\alpha}{2} t$

$\large x \propto t^2$

## Two trains one travelling at 54 kmph and the other at 72 kmph are headed towards one another along a straight track…

Q: Two trains one travelling at 54 kmph and the other at 72 kmph are headed towards one another along a straight track. When they are 1/2 km apart, both derives simultaneously see the other train and apply their brakes. If each train is decelerated at the rate of 1 m-s2, will there be a collision ?

Sol. Velocity of the first train is 54 kmph = 15 m/s Distance travelled by the first train before coming to rest

$\large S_1 = \frac{u^2}{2a} = \frac{225}{2 \times 1}$

S1 = 112.5 m

Velocity of the second train is 72 kmph = 20 m/s

Distance travelled by the second train before coming to rest

$\large S_2 = \frac{u^2}{2a} = \frac{400}{2 \times 1}$

S2 = 200 m

Total distance travelled by the two trains before coming to rest

S = S1 + S2

= 112.5 + 200 = 312.5 m

Because the greater than 312.5 m, there will be no collision between the two trains.

## A car is moving with a velocity of 40 m/s. the drivers sees a stationary truck ahead at a distance of 200m…

Q: A car is moving with a velocity of 40 m/s. the drivers sees a stationary truck ahead at a distance of 200 m. after some reaction time Δt the breaks are applied producing a (reaction) retardation of 8 m/s2. What is the maximum reaction time to avoid collision ?

Sol. The distance travelled by the car is 100 m

To avoid the clash, the remaining distance 200 – 100 = 100 m must be covered by the car with uniform velocity 40 m/s during the reaction time Δt.

100/(Δt) = 40

⇒ Δt = 2.5 s

The maximum reaction time Δt = 2.5 s

## A car start from rest and moves with uniform acceleration of 5 m/s2 for 8 sec. if the acceleration ceases after 8 seconds…

Q: A car start from rest and moves with uniform acceleration of 5 m/s2 for 8 sec. if the acceleration ceases after 8 seconds then find the distance covered in 12 s starting from rest.

Sol. The velocity after 8 sec v = 0 + 5 × 8 = 40 m/s

Distance covered in 8 sec

S = 0 + (1/2) × 5 × 64 = 160 m

After 8s the body moves with uniform velocity and distance covered in next 4s with uniform velocity.

S = vt = 40 × 4 = 160 m

The distance covered in 12 s = 160 + 160 = 320 m.