Assume that each of the follwing functions gives the position of an object at time t. Find the velocities indicated by setting up and evaluating a limt algebraically.
(a) s(t) = 3t^2, Find v(-1).
(b) s(t) = 1/t , Find v(2).
(c) s(t) = square root of t, Find v(1).
The answers that I got were the following...
a)3
b)1/2
c)1
I did them another way and the answers I got were
a)-6
b) -(1/4)
c) I didn't really get any answers here.
Please tell me which answers are right or if any of them are right...
To find the velocity of an object at a specific time, we can calculate the derivative of the given position function with respect to time. In other words, we need to find the derivative of the function s(t).
Let's go through each part and find the correct velocities:
(a) s(t) = 3t^2, find v(-1):
To find v(-1), we need to calculate the derivative of s(t) first:
s'(t) = d/dt(3t^2) = 6t
Now, we substitute t = -1 into the derivative to find v(-1):
v(-1) = 6(-1) = -6
So, the correct answer for part (a) is -6.
(b) s(t) = 1/t, find v(2):
Similarly, we find the derivative of s(t) with respect to t:
s'(t) = d/dt(1/t) = -1/t^2
Now, substitute t = 2 into the derivative to calculate v(2):
v(2) = -1/(2^2) = -1/4
So, the correct answer for part (b) is -1/4.
(c) s(t) = √t, find v(1):
Again, take the derivative of s(t) with respect to t:
s'(t) = d/dt(√t) = 1/(2√t)
Then, substitute t = 1 into the derivative to find v(1):
v(1) = 1/(2√1) = 1/2
So, the correct answer for part (c) is 1/2.
Therefore, the correct answers are:
(a) v(-1) = -6
(b) v(2) = -1/4
(c) v(1) = 1/2
Your first set of answers is correct, while the second set of answers is incorrect.