Answer:
0.78m/s
Explanation:
We are given that
Acceleration=[tex]a=\frac{1}{4}s^{\frac{1}{2}}m/s^2[/tex]
v=0, s=1 when t=0
We have to find the particle's velocity at s=2m
We know that
[tex]a=\frac{dv}{dt}=\frac{dv}{ds}\times \frac{ds}{dt}=\frac{dv}{ds}v[/tex]
[tex]vdv=ads[/tex]
[tex]\int_{0}^{v} vdv=\int_{1}^{s}0.25s^{\frac{1}{2}}ds[/tex]
[tex]\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2})^{s}_{1}[/tex]
By using formula:[tex]\int x^ndx=\frac{x^{n+1}}{n+1}+C[/tex]
[tex]\frac{v^2}{2}=0.25\times \frac{2}{3}(s^{\frac{3}{2}}-1[/tex]
Substitute s=2
[tex]\frac{v^2}{2}=\frac{0.50}{3}((2^{1.5})-1)[/tex]
[tex]\frac{v^2}{2}=\frac{0.50}{3}\times 1.83[/tex]
[tex]v^2=2\times 0.305=0.61[/tex]
[tex]v=\sqrt{0.61}=0.78m/s[/tex]
Hence, the velocity of particle at s=2m=0.78m/s
