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

0.78m/s

Step-by-step explanation:

We are given that

Acceleration=
`a=(1)/(4)s^{(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=(dv)/(dt)=(dv)/(ds)* (ds)/(dt)=(dv)/(ds)v`

`vdv=ads`

`\int_(0)^(v) vdv=\int_(1)^(s)0.25s^{(1)/(2)}ds`

`(v^2)/(2)=0.25* (2)/(3)(s^{(3)/(2))^(s)_(1)`

By using formula:
`\int x^ndx=(x^(n+1))/(n+1)+C`

`(v^2)/(2)=0.25* (2)/(3)(s^{(3)/(2)}-1`

Substitute s=2

`(v^2)/(2)=(0.50)/(3)((2^(1.5))-1)`

`(v^2)/(2)=(0.50)/(3)* 1.83`

`v^2=2* 0.305=0.61`

`v=√(0.61)=0.78m/s`

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