Sure! I can help you with that.
To find the velocity indicated by setting up and evaluating a limit algebraically, you need to take the derivative of the position function with respect to time.
Given the position function: s(t) = √t
Step 1: Find the derivative of the position function, which represents the instantaneous velocity:
s'(t) = d/dt (√t)
Step 2: Apply the power rule for differentiation:
s'(t) = (1/2) * t^(-1/2)
Step 3: Now, we want to find the velocity at t = 1, which is represented by v(1).
To evaluate this limit algebraically, we set up the limit expression:
v(1) = lim (t->1) [s'(t)]
Step 4: Substitute the derivative function into the limit expression:
v(1) = lim (t->1) [(1/2) * t^(-1/2)]
Step 5: Evaluate the limit by substituting the value of t:
v(1) = (1/2) * (1)^(-1/2)
Now, to simplify this expression, recall that (1)^(-1/2) is equal to 1 divided by the square root of 1, which is 1/1 = 1.
v(1) = (1/2) * 1
Therefore, the velocity at t = 1 is v(1) = 1/2.
So, the answer you provided, 1/2, is indeed the correct velocity when evaluating the limit algebraically.