Question

A particle moves on a straight line so that its velocity at time t is given by v = sqrt(s)
where s is the distance from the origin.

s=1 when t=0, then what is the value of s when t=1 ?

I tried using parameters but I couldn't find a good way ... can someone help ?

Answer 1

I would really love to have some help ... :(

Answer 2

velocity is rate of distance/time
v=ds/dt = sqrt(s)
rearrange variables so s is with ds
ds/sqrt(s) = dt
now integrate both sides
integral 1/sqrt(s) = integral s^-1/2 = 2*s^1/2 = 2sqrt(s)
2sqrt(s) = t + C
solve for s
sqrt(s) = t/2 + C
s = (t/2 +C)^2 = t^2/4 +Ct +C^2
find C using initial conditions
s=1 when t=0
1=C^2 -->C=1
s = t^2/4 + t + 1
when t=1
s = 1/4+1+1 = 2.25

Hope this helps