Question

how to find instantaneous velocity with the function s(t)=4t+1 at t=2 seconds

Answer 1

its velocity is not changing and it is always 4m/s

Answer 2

So, I believe instantaneous velocity is the first derivative of the function.

\[s \prime (t) = 4\] Meaning that there is no place to plug in t=2.

So instantaneous velocity is static and should be 4.

Answer 3

where did you get the 4?

Answer 4

differenciate s(t) wrt time

Answer 5

You need to find the first derivative of of s(t)= 4t+1.

The derivative of 4t is 4(1) and the derivative of 1 is 0.

s'(t)= 4(1)+0 = 4

Answer 6

But doesn't the problem strep that t=2?

Answer 7

Also if the function is 4t+1 shouldn't plugging 1 in be 5?

Answer 8

No, because when you are looking for instantaneous velocity you need to plug t into the derivative of the original equation. However, in this case in the derivative equation there is no longer a t.

Answer 9

Basically, you were given s(t)=4t+1 at t=2.

The next step is to find the derivative of s(t)= 4t+1.
s'(t) = 4

Now you plug t=2 into s'(t) = 4.

However, there is no longer a t to plug into so the answer is just 4.

Answer 10

Ok, so how did you find the derivative? did you use the slope equation?

Answer 11

No, I did it in a way you might not have learned yet. But let me know if I'm wrong. Using the difference quotient:

\[\lim_{h \rightarrow 0} \frac{ (4(t-h)+1)-(4t+1) }{ h }\]

can you solve for the derivative from there?

Answer 12

The way I did it was knowing these rules:

d/dx (f(x)+g(x)) = f'(x)+g'(x) (derivative of sums is sum of derivative)

d/dx cf(x) = cf'(x)
(when taking the derivative of a constant, take out the constant and put it back after finding the derivative of f(x).

d/dx x = 1 (derivative of a variable by itself is 1)

d/dc c = 0 (derivative of a constant is 0)

Answer 13

In the last line the equation should be d/dx c= 0. Sorry for the typo.

Hope I helped clarify a bit :)

Answer 14

Yeah, I didn't learn that, thanks anyway!