How Far Does a Runner Travel with Varying Speeds?

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Homework Help Overview

The problem involves a runner's motion characterized by acceleration, constant speed, and deceleration, with the goal of determining the total distance covered over a 20-second period.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss calculations for distance during different phases of motion, including acceleration, constant speed, and deceleration. Questions arise regarding the assumptions of constant velocity during acceleration and deceleration phases.

Discussion Status

Some participants have provided calculations for distance covered in each phase, while others question the validity of these calculations based on the nature of acceleration and deceleration. There is an exploration of using a velocity-time graph as a potential approach to clarify the problem.

Contextual Notes

Participants note the assumption of constant acceleration and deceleration, which may not hold true in the context of the problem. There is also mention of varying interpretations of average velocity during these 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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