An object moves along a coordinate line with acceleration a(t)=(t+1)^-1/2 units/second.
a. Find velocity function given that initial velocity is 2 units/second.
b. Find the position that the initial velocity is 1 unit/second and the initial position of the object is 0.
For part A I understand that a(t)=v'(t) and I need to differentiate. Therefore I found that v(t)=2(t-1)^1/2 + c
However, I don't know if this is right, where does the vo+at come in? And what about the initial velocity of 2 units/second?
For both parts would appreciate steps as to how to go about it, not just looking for the answer. Looking for feedback. Thank you in advance.
acceleration means the differentiation of velocity.
`(dv(t))/dt = a(t)`
To get the velocity function we need to integrate the function.
`dv(t) = a(t)*dt`
`intdv(t) = int a(t)dt`
`v(t) = int a(t)dt + C` where C is a constant.
`= int a(t)dt + C`
`= int (t+1)^(-1/2)dt + C`
`= (t+1)^(-1/2+1)/(-1/2+1) + C`
`= 2(t+1)^(1/2) + C`
It is given that initially the velocity is 2. That means at time t = 0 the velocity is 2 or v(0) = 2
`v(t) = 2(t+1)^(1/2) + C`
`v(0) = 2(0+1)^(1/2) + C`
`2 = 2+C`
`C = 0`
Velocity function `v(t) = 2(t+1)^(1/2)`
you can use `(ds(t))/dt = v(t)` combination to get the position function of the object. Use the same procedure as the velocity function.