How Far Will You Travel Before Catching Up with the Biker?

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SUMMARY

The problem involves calculating the distance at which a car, starting from rest and accelerating at 3 m/s², will catch up to a biker traveling at a constant speed of 9 m/s. The equations of motion for both the car and the biker must be equated to find the time and distance at which they meet. The biker's displacement is linear, while the car's displacement is quadratic due to its acceleration. The solution requires setting the areas under the velocity-time graphs equal to each other to determine the catch-up point.

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Need help with "motion" question..please!

Homework Statement


you are sitting in a car that is stopped. The moment a biker passes you, you start accelerating at 3 m/s2, while the bicycle continues to move along at a constant speed of 9 m/s. After what distance will you catch up with the biker?

From the question, the info I have is v=9m/s V0=0 a=3m/s*2



So I think I would put it into V=V0+at and solve for t. Then use all info including the new t into x=x0+v0+1/2at*2

I keep getting it wrong...any ideas?
 
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suppose after tsec , the biker is caught.
equate the displacements , and find necesary condition for quadratic to be real .
 


Draw the v-t diagram for your motion and the bikers motion.
Your motion is a triangle, the bikers is flat.
You need the time when the areas under the graphs (you know how to find the area of a triangle right?) are equal.

You could also find the equation for the distance between the bike and the car at time t, then find the time that this distance is zero.
If the bike goes at a constant speed v - what is the equation for it's displacement with time?
If the car has a uniform acceleration of a, from rest, what is the equation for it's displacement with time?
Subtract these two.

Off what you did:
So I think I would put it into V=V0+at and solve for t. Then use all info including the new t into x=x0+v0+1/2at*2
... presumably for the car: what did you use for the final speed of the car? (note: it will be going much faster than the bike when it catches up.)
 
Last edited:

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