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?
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!
To find the position x(t) at time t, you need to integrate the velocity function x'(t) with respect to t. However, it seems you made a mistake in your integration step.
Given: x'(t) = 1/((2t + 1)^(1/2)), x(0) = 0
To find x(t), let's integrate x'(t) with respect to t:
∫x'(t) dt = ∫(1/((2t + 1)^(1/2))) dt
To simplify this integral, we can use a substitution. Let u = 2t + 1. Then, du/dt = 2, and dt = du/2:
∫(1/((2t + 1)^(1/2))) dt = ∫(1/u^(1/2)) dt = ∫(1/u^(1/2)) (du/2)
Now the integral becomes:
(1/2) ∫(u^(-1/2)) du
Using the power rule for integration, the integral evaluates to:
(1/2) * (2(u^(1/2))) + C
Simplifying further, we get:
u^(1/2) + C' = (2t + 1)^(1/2) + C'
Since x(0) = 0, we can solve for C' by substituting t = 0 and x = 0 into this equation:
0 = (2(0) + 1)^(1/2) + C'
0 = 1^(1/2) + C'
0 = 1 + C'
C' = -1
Therefore, the position x(t) at time t should be:
x(t) = (2t + 1)^(1/2) - 1
Now let's solve for the time when the object is at x = 3:
3 = (2t + 1)^(1/2) - 1
To isolate the term (2t + 1)^(1/2), add 1 to both sides:
4 = (2t + 1)^(1/2)
Next, square both sides to get rid of the square root:
16 = 2t + 1
Subtract 1 from both sides:
15 = 2t
Divide both sides by 2:
t = 15/2
So the correct answer is t = 15/2.
To find the position x(t) at time t, we can integrate the velocity function v(t) = x'(t).
Given that x'(t) = 1/((2t + 1)^(1/2)) and x(0) = 0, we can integrate x'(t) with respect to t to find x(t):
∫ x'(t) dt = ∫ (1/((2t + 1)^(1/2))) dt
To evaluate the integral, we can use the substitution u = 2t + 1, which gives du = 2dt. Rearranging the equation, we have dt = (1/2)du. Substituting these values, we get:
∫ (1/((2t + 1)^(1/2))) dt = ∫ (1/((2t + 1)^(1/2))) (1/2)du
= (1/2) ∫ (1/u^(1/2)) du
= (1/2) ∫ u^(-1/2) du
= (1/2) * 2u^(1/2) + C
= u^(1/2) + C
= (2t + 1)^(1/2) + C.
Now, to determine the integration constant C, we use the initial condition x(0) = 0. Substituting t = 0 into our expression for x(t), we have:
x(0) = (2(0) + 1)^(1/2) + C
0 = 1^(1/2) + C
0 = 1 + C
C = -1.
Therefore, the position x(t) at time t is:
x(t) = (2t + 1)^(1/2) - 1.
Now, let's consider finding the time when the object is at x = 3. We can substitute x = 3 into our expression for x(t) and solve for t:
3 = (2t + 1)^(1/2) - 1.
To isolate the square root, we can add 1 to both sides of the equation:
4 = (2t + 1)^(1/2).
Now, we square both sides of the equation:
16 = 2t + 1.
Rearranging the equation and solving for t, we have:
2t = 16 - 1
2t = 15
t = 15/2.
Therefore, the time when the object is at x = 3 is t = 15/2. You initially obtained t = 1/2 because you missed the step of squaring both sides of the equation when isolating the square root.