# Distance covered by a runner

1. Sep 20, 2007

### rgluckin

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

2. Sep 20, 2007

### darklide

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

3. Sep 20, 2007

### rgluckin

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?

4. Sep 20, 2007

### darklide

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)