Question

A particle is moving on a straight line with velocity v=sqrt(s) where s is the distance from the particle to the origin. When t=0, s =1. What is the value of s when t=1 ?

Answer 1

v=ds/dt
ds/dt=s^1/2
s^-1/2 ds = dt integrate both sides
2*s^1/2=t+c solve for s
s^1/2=(t+c)/2
s=((t+c)/2)^2
s=(t^2+2tc+c^2)/4 substitute s(t=0)=1 to find c
1=0+0+(c^2)/4
c=2
full solution: s=(t^2+4t+4)/4
substitute t=1
s=(1+4+4)/4=9/4
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Answer 2

That makes so much sense.
So this is a problem that has to do with differential equations, huh?
I thought it had something to do with parametric equations, so I couldn't find out how to solve it...
Thanks !