Question

The acceleration of a particle traveling along a straight line isa=14s1/2m/s2, wheresis in meters. Ifv= 0,s= 1 m whent= 0, determine the particle’svelocity ats= 2 m.

Answer 1

Answer:

0.78m/s

Explanation:

We are given that

Acceleration=$a=\frac{1}{4}s^{\frac{1}{2}}m/s^2$

v=0, s=1 when t=0

We have to find the particle's velocity at s=2m

We know that

$a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v$

$vdv=ads$

$\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}$

By using formula:$\int x^ndx=\frac{x^{n+1}}{n+1}+C$

$\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1$

Substitute s=2

$\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)$

$\frac{v^2}{2}=\frac{0.50}{3}\times 1.83$

$v^2=2\times 0.305=0.61$

$v=\sqrt{0.61}=0.78m/s$

Hence, the velocity of particle at s=2m=0.78m/s