Graphical Analysis of One-Dimensional Motion

Question

A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See figure below. For each answer, enter a number.)

A graph of Runner Velocity vs. Time has a horizontal axis labeled Time (s) and a vertical axis labeled Velocity (m/s). Straight lines connect adjacent points, which have approximate values as follows.

1. (0, 0), 
2. (2, 6), 
3. (4, 12), 
4. (6, 12), 
5. (8, 12), 
6. and (10, 12).

(a)

What is his average velocity (in m/s) for the first 4 s?

 m/s

(b)

What is his instantaneous velocity (in m/s) at t = 6 s?

 m/s

(c)

What is his average acceleration (in m/s²) between 0 and 4 s?

 m/s²

(d)

What is his time (in s) for the race?

 s

Answer 1

a) You can read it off the graph or use v_avg = (v₂+v₁)/2 for the interval t= 0 to 4 and const a (linear)

b) Instantaneous velocity is read off of the graph at t = 6.

c) avg a = (v₂-v₁)/(t₂-t₁) for t = 0 to 4 (avg a is the total velocity change divided by total time interval)

d) This is a little deceptive as the d under the curve is 1/2(12 m/s)(4s) + 12m/s(10s - 4 s) = 96 meters, which is not the length of the race. I think this is probably a mistake and the answer they are looking for is 10 seconds. If not you need to add 4 meters/12 m/s to the 10 seconds.

Please consider a tutor.