# Thread: Find the Average Velocity

1. ## Find the Average Velocity

The position function s of a point P moving on a coordinate line l is given, with t in seconds and s(t) in centimeters.

(a) Find the average velocity of P in the following interval [1, 1.2].

(b) Find the velocity of P at t = 1.

s(t) = 4t^2 + 3t

2. ## Re: Find the Average Velocity

a) Average velocity is $\displaystyle v_{avg} = \frac {\Delta s}{\Delta t}$. So determine $\displaystyle s(1.2)-s(1.0)$ and divide by $\displaystyle \Delta t = 0.2$ seconds.

b) Given the position function s(t), the velocity equations is $\displaystyle v(t) = \frac {ds(t)}{dt}$. Evaluate for t=1.0.

3. ## Re: Find the Average Velocity

a) I got average velocity = 11.8

b) I am confused to do it.

4. ## Re: Find the Average Velocity

Originally Posted by joshuaa
a) I got average velocity = 11.8
Good!

Originally Posted by joshuaa
b) I am confused to do it.
The velocity of the particle at t=1 is equal to the slope of the s(t) curve at t=1. I suggested calculating the derivative to find that slope - have you taken any calculus classes where you have learned about derivatives? If not, then you could plot the graph of s=4^2+3t and use a straight edge to estimate its slope at t=1. Alternatively, you could use the same approach as in part (a) to find the average velocity between two points that are very close to t=1; for example if you calculate the average velocity over the interval t= (0.99, 1.01) you will get a very good approximation of the actual velocity at t=1.

5. ## Re: Find the Average Velocity

Thank you ebaines for the detailed explanation. Yes, I know how to find the slope using the Derivative.

When I find the derivative of 4t^2 + 3t, I get 8t + 3.

If my calculations went correctly, the Velocity at t = 1, might be 8(1) + 3 = 11.

Correct!