# 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...

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.

*print*Print*list*Cite

### 1 Answer

`a(t) =(t+1)^(-1/2)`

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.

`v(t) `

`= 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)`

Note:

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.