# Distance covered by a runner

by rgluckin
Tags: covered, distance, runner
 P: 2 1. The problem statement, all variables and given/known data 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? 2. Relevant equations 3. 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. 1. The problem statement, all variables and given/known data 2. Relevant equations 3. The attempt at a solution
 P: 13 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
 P: 2 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?
P: 13

## Distance covered by a runner

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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