The speed of a runner increased steadily during the first three seconds of a race. Her speed at half-second intervals is given in the table. Find a lower estimate for the distance that she traveled during these three seconds.
t(s) 0 0.5 1.0 1.5 2.0 2.5 3.0
v(ft/s) 0 3 6.6 9.9 11.5 11.8 12
I know that delta x is 3/7 and I tried multiplying t*v and then doing the area thing... I don't know what to do please help!
low estimate= avg veloctity*time= 12/2 *3= 18m
The area thing ought to give you the actual distance.
i got it thanks!
You're welcome! I'm glad you were able to find the answer. Just to clarify, the lower estimate for the distance traveled by the runner during the first three seconds is 18 feet.
To arrive at this estimate, you used the formula for average velocity, which is distance divided by time. In this case, the average velocity is calculated as (12 ft/s + 0 ft/s) / 2 * 3 s = 18 ft.
If you wanted a more accurate measure of the distance, you can use the area under the velocity-time graph. Since the speed of the runner increased steadily, you can approximate the area between each pair of points with a trapezoid. The sum of these trapezoid areas would give you a more precise estimate for the distance traveled.