Question

Q2/Deceleration of a particle is based on relation a=-3 v² m/s² where v in m/s. If it moves along a straight line and has velocity 10 m/s and position s = 8m when t=0, determine its velocity and position when t= 3 s. Where the velocity become zero. Discuss briefly.​

Answer 1

Explanation:

Given: a = -3v^2

By definition, the acceleration is the time derivative of velocity v:

$a = \frac{dv}{dt} = - 3 {v}^{2}$

Re-arranging the expression above, we get

$\frac{dv}{ {v}^{2} } = - 3dt$

Integrating this expression, we get

$\int \frac{dv}{ {v}^{2} } = \int {v}^{ - 2}dv = - 3\int dt$

$- \frac{1}{v} = - 3t + k$

Since v = 10 when t = 0, that gives us k = -1/10. The expression for v can then be written as

$- \frac{1}{v} = - 3t - \frac{1}{10} = - ( \frac{30 + 1}{10} )$

or

$v = \frac{10}{30t +1 }$

We also know that

$v = \frac{ds}{dt}$

or

$ds = vdt = \frac{10 \: dt}{30t + 1}$

We can integrate this to get s:

$s = \int v \: dt = \int ( \frac{10}{30t + 1}) \: dt = 10 \int \frac{dt}{30t + 1}$

Let u = 30t +1

du = 30dt

so

$\int \frac{dt}{30t + 1} = \frac{1}{30} \int \frac{du}{u} = \frac{1}{30}\ln |u| + k$

$= \frac{1}{30}\ln |30t + 1| + k$

So we can now write s as

$s = \frac{1}{3}\ln |30t + 1| + k$

We know that when t = 0, s = 8 m, therefore k = 8 m.

$s = \frac{1}{3}\ln |30t + 1| + 8$

Next, we need to find the position and velocity at t = 3 s. At t = 3 s,

$v = \frac{10}{30(3) +1 } = \frac{10}{91}\frac{m}{s} = 0.11 \: \frac{m}{s}$

$s = \frac{1}{3}\ln |30(3) + 1| + 8 = 9.5 \: m$

Note: velocity approaches zero as t --> $\infty$