Question

The function s = f(t) gives the position of a body moving on a coordinate line, with s in meters and t in seconds.
s = 3t^2 + 4t + 3, 0 <= t <= 2
Find the body's speed and acceleration at the end of the time interval.

Answer 1

\[\large f(t)=s, \qquad\text{position}\]\[\large f'(t)=v,\qquad\text{velocity}\]\[\large f''(t)=a, \qquad\text{acceleration}\]

In this problem they want us to find,\[\large f'(2)\qquad \text{and}\qquad f''(2)\]

Do you understand how to take the derivative of the function, using the power rule?

Answer 2

Or umm I should be a little more clear on one of those points.
They want `speed`, not velocity.

Velocity would include a direction, such as negative or positive.
Speed is a scalar, no direction.

So whatever value we get for \(\large f'(2)\), we take the absolute value of it.

Answer 3

Ah, alright. So f'(t) = 6t + 4, and substituting 2 in you would get 6(2) + 4 which equals 16, so its speed at two seconds would be 16 m/sec? But wouldn't f'' (t) just equal 6?

Answer 4

The position function is quadratric,
The velocity funtion (which gives us speed) is linear,
So the acceleration will be constant.

So the acceleration will be the same value at every point.
Yah 6 that sounds correct! :) Good job!

Answer 5

Awesome! Thank you very much for the help!