Find the instantaneous velocity, using limits, for the given value of t of an object moving according the given equation.
a. s(t)=t^2–8t+18 ; t=3
b. 𝑠(𝑡)=1/t^2 ; t=1
v(t) = ds/dt, so
(a) v(t) = 2t-8
(b) v(t) = -2/t^3
Just plug in your numbers.
steve i got that for a but we werent allowed to use derivative rules to get the answer because we havent learned those yet. and from getting 2t-8 would i just plug in 3 for t because if so i got -2
yes.
I forgot that it said "using limits". Sorry. It's just back to basics.
Thanks for your help Steve. I appreciate it. and yes it is back to basic tedious work where as doing it using derivatation is a lot less steps
To find the instantaneous velocity using limits for a given value of t, we need to take the derivative of the position function with respect to time.
a. s(t) = t^2 - 8t + 18
To find the instantaneous velocity at t = 3, we need to find the derivative of s(t) and evaluate it at t = 3.
Step 1: Find the derivative of s(t)
The derivative of t^2 is 2t.
The derivative of -8t is -8.
The derivative of 18 is 0 (as it's a constant).
Therefore, the derivative of s(t) is ds/dt = 2t - 8.
Step 2: Evaluate the derivative at t = 3
Substitute t = 3 into the derivative expression:
ds/dt = 2(3) - 8
ds/dt = 6 - 8
ds/dt = -2
So, the instantaneous velocity of the object at t = 3 is -2 units per time.
b. s(t) = 1/t^2
To find the instantaneous velocity at t = 1, we again need to find the derivative of s(t) and evaluate it at t = 1.
Step 1: Find the derivative of s(t)
Using the power rule, the derivative of 1/t^2 is -2/t^3.
Step 2: Evaluate the derivative at t = 1
Substitute t = 1 into the derivative expression:
ds/dt = -2/(1^3)
ds/dt = -2/1
ds/dt = -2
The instantaneous velocity of the object at t = 1 is -2 units per time.