Solving Kinematics Problem: Who Wins the Race?

AI Thread Summary
Alvin has a head start of 13 meters and runs at a constant speed of 7 m/s, completing the race in approximately 12.4 seconds. Ophelia starts at 1.2 m/s and accelerates at 1.5 m/s², finishing the race in about 10.8 seconds. To determine when Ophelia catches up, the equations for their displacements are set equal, leading to a quadratic equation. Solving this gives the time of catch-up as approximately 9.55 seconds, with both racers at about 79.84 meters. The calculations confirm that Ophelia wins the race despite her slower initial speed.
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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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