The indefinite integral is
x(t)= (2t + 1)^{1/2) + C
Since the value is 0 when t = 0, C = -1.
x(t)= (2t + 1)^{1/2) -1
You are OK so far
When t = 3,
4 = (2t + 1)^(0.5)
16 = 2t +1
2t = 15
t = 15/2
You did the calculus great but not the easy part!
The velocity v(t) = x'(t) at time t of an object moving along the x axis is given, along with the initial position x(0) of the object. In each case, find:
5) x'(t) = 1/((2t + 1)^(1/2)); x(0) = 0
a) the position x(t) at time t
I found that x(t) = (2t + 1)^{1/2) - 1
c) The time when the object is at x = 3
3 = (2t + 1)^(0.5) - 1
4 = (2t + 1)^(0.5)
2 = 2t + 1
1/2 = t
...but the answer is t = 15/2, what did I do wrong?
1 answer