How Far Does a Runner Travel with Varying Speeds?

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A runner accelerates from 0 to 15 ft/sec in 2 seconds, maintains that speed for 8 seconds, and then decelerates to 0 ft/sec over 10 seconds. The total distance covered during acceleration is approximately 8 feet, while the distance at constant speed is 120 feet. During deceleration, the average speed is calculated as 7.5 ft/sec, resulting in an additional 75 feet. The total distance covered over the 20 seconds is 203 feet. The discussion highlights the importance of understanding average velocity during acceleration and deceleration phases.
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Homework Statement



A runner accelerates from 0 to 15 ft/sec in 2 seconds.
She goes at a constant v from 2 sec to 10 sec.
She deccelerates to 0 at the 20 sec mark

Q: How much distance does she cover?

Homework Equations





The Attempt at a Solution



Here is my work:
From 0 to 2 sec, she accelerates from 0 ft/sec to 15 ft/sec.
Her average velocity is 7.5 ft/sec, so in 2 sec she covers 7.5 ft. (round to 8)
From 2 sec to 10 sec, she goes at a constant velocity of 15 ft/sec, so the distance covered is 15ft/sec*8sec, which is 120 feet
From 10 sec to 20 sec, she decelerates from 15 ft/sec to 0 ft/sec
Avg velocity of 15ft/sec / 10 sec = 1.5 ft/sec * 10 sec = 15 ft

8 ft + 120ft + 15ft = 143 ft in 20 sec.
 
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D =(0.5*2*15)+(8*15)+(0.5*10*15)
=15+120+75
=180ft in 20 sec assuming that there is constant acc and deceleration
 
OK, I get the first two figures -- but for the decceleration, she goes from 15ft/sec to 0, for a difference of 15 in 10 seconds, so that's an avg velocity of 1.5ft/sec *10 sec is 15 ft.

Where did I go wrong?
 
You are assuming that the velocity is constant(average) which is not the case when accelerating or decelerating.
I did the question by using a velocity-time graph.
U could try to do so...
(i may be wrong...havent done these kind of questions for nearly one year)
 
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