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...
a) -6
b) -1/4
c) 1/2
a) -6
b) - 1/4
c) 1/2 = square root of t is
1/2(square root of t)
Thank you very much for the clarification, could you explain to me how you got 1/2 in c please??
s(t) = √t = t(1/2)
s '(t) = (1/2)t^(-1/2) = 1/(2√t)
s '(1) = 1/(2√1) = 1/2
Could you please tell me how to do it evaluating the limt algebraically? I'm trying to do it and I can't get it....
To find the velocity of an object at a specific time t using the given position function, we differentiate the position function with respect to time.
(a) For the position function s(t) = 3t^2, we need to find v(-1). To do this, we differentiate s(t) with respect to t.
ds/dt = 6t
Now, substitute t = -1 into the derivative to find the velocity:
v(-1) = 6(-1) = -6
So, the correct answer is -6.
(b) For the position function s(t) = 1/t, we need to find v(2). Differentiate s(t) with respect to t:
ds/dt = -1/t^2
Now, substitute t = 2 into the derivative to find the velocity:
v(2) = -1/2^2 = -1/4
So, the correct answer is -1/4.
(c) For the position function s(t) = √(t), we need to find v(1). Differentiate s(t) with respect to t:
ds/dt = 1/(2√t)
Now, substitute t = 1 into the derivative to find the velocity:
v(1) = 1/(2√1) = 1/2
So, the correct answer is 1/2.
Based on the given position functions and the differentiation process described, it seems like the answers you've obtained using the other method are incorrect. The correct answers are:
(a) v(-1) = -6
(b) v(2) = -1/4
(c) v(1) = 1/2