Solving Kinematics Problem: Who Wins the Race?

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Homework Help Overview

The problem involves a kinematics scenario where two individuals, Alvin and Ophelia, race to a classroom. Alvin has a head start and travels at a constant speed, while Ophelia starts slower but accelerates. The goal is to determine who wins the race and when Ophelia catches up to Alvin.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the setup of the equations of motion for both racers and how to equate their displacements to find when they meet. There are attempts to apply the quadratic formula to solve for time, with some participants questioning the setup of the equations.

Discussion Status

The discussion is ongoing, with some participants providing guidance on how to set up the equations correctly. There is acknowledgment of the original poster's attempts and some clarification on the approach needed to find the time and distance at which Ophelia catches up to Alvin.

Contextual Notes

Participants note the importance of correctly applying the equations of motion and the implications of the initial conditions, such as Alvin's head start and Ophelia's acceleration. There is also mention of potential errors in the quadratic setup that need to be addressed.

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Homework Statement



Alvin races Ophelia to Physics class. Alvin has a headstar of 13m and travels at a constant speed of 7m/s. Phelia is initially traveling at 1.2m/s but then begins to accelerate at 1.5m/s2 until she reaches the physics classroom 100m away from her.

Who wins the race? when and where did ophelia catch up? (both metres and time)

Homework Equations



d=vit+1/2(a)(t)2+di

The Attempt at a Solution


Who wins the race? I was able to figure out that alvin completed the race at 12.4s and phelia competed the race at 10.8s (though I'm not sure if its right)

I got stuck trying to find out WHEN they caught up. I tried setting the equation to

vit+1/2(a)(t)2+di=vit+1/2(a)(t)2+di

and plucking in the numbers for both sides but when I tried to find the variable for time, I put it into the quadratics formula. I ended up have no real roots (square rooting negatives)
 
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Basically you want to find the time when Phelia's displacement equals Alvin's displacement plus 13 meters. Or \triangle D_P = \triangle D_A + 13
 
Last edited:
If you know where they crossed, plug that x value into Alvin's EOM to get t.
 
Start by listing the information you have:

df = 100 m

Alvin

di = 13 m

v(t) = vi = 7 m/s

a = 0

==> d(t) = di + vit = 13 + 7t

Ophelia

di = 0 m

v(t) = vi = 1.2 m/s

a = 1.5 m/s2

==> d(t) = vit + (1/2)at2 = 1.2t + 0.75t2

It seems that your approach for answering the first question is correct: find out how long it takes for each person to run 100 m:

Alvin:

100 = 13 + 7t

t = (87 m) / (7 m/s) = 12.43 s

Ophelia:

100 = 1.2t + 0.75t2 =

(3/4)t2 + (6/5)t - 100 = 0

using the quadratic formula, t = 10.77 s

It looks like you're doing alright so far.

Now, to find the the position and time at which Ophelia catches up, you equate the two trajectories (since their positions are equal at this instant)

13 + 7t = (3/4)t2 + (6/5)t

0.75t2 - 5.8t - 13 = 0

Solve for t using the quadratic formula with

a = 3/4

b = -5.8 = -29/5

c = -13

t = 9.5486 s

Alvin:

d(t) = d(9.55) = 13 + 7(9.55) = 79.8402 m

equivalently

Ophelia

d(9.55) = 1.2(9.55) + 0.75(9.55)2 = 79.8402 m

Since your methods are right, I went over the full solution. You must have just made a mistake in setting up the final quadratic equation
 

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