Time when two cars have equal speeds

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The discussion revolves around determining when two toy cars have equal speeds given their initial conditions and accelerations. The first car starts with an initial velocity of -3 cm/s and accelerates at 2.90 cm/s², while the second car moves at a constant velocity of 6.30 cm/s. Initial calculations suggested the time for equal speeds was 1.1379 seconds, but the correct answer is 3.21 seconds, with equal speeds of 12.309 cm/s. The confusion arises from the understanding of the first car's negative initial velocity and its need to reverse direction before matching speeds with the second car. Accurate calculations must account for the direction and magnitude of velocities to find the correct time of equal speeds.
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Homework Statement



At t = 0, one toy car is set rolling on a straight track with initial position 16.0 cm, initial velocity -3 cm/s, and constant acceleration 2.90 cm/s2. At the same moment, another toy car is set rolling on an adjacent track with initial position 10.5 cm, an initial velocity of 6.30 cm/s, and constant zero acceleration.
(a) At what time, if any, do the two cars have equal speeds?


The Attempt at a Solution



My answer is 1.1379s by using the method below,
Vf = Vi +at
6.3 = 3 + (2.9)t
t = 1.1379s

However, the answer is 3.21s, with the equal speeds as 12.309cm/s(I suppose by using Vf = Vi +at).

I don't understand why the final common speed is not 6.30 cm/s since the one of the object(with initial velocity of 6.30 cm/s) is moving under zero acceleration. Shouldn't it moves under a constant velocity throughout the whole journey?
 
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The initial velocity of the first car is -3cm/s: vf=-3+2.9t. The question is when the cars have the same speed, that is the same magnitude of their velocity. This means either v1=v2 or v1=-v2. ehild

i
 
ehild said:
The initial velocity of the first car is -3cm/s: vf=-3+2.9t. The question is when the cars have the same speed, that is the same magnitude of their velocity. This means either v1=v2 or v1=-v2.


ehild

i

yes, I know this clearly. unfortunately, I couldn't get the correct answer.
 
MechaMZ said:
yes, I know this clearly. unfortunately, I couldn't get the correct answer.
Can you show your re-work when you use the correct initial velocity, -3 cm/s?
 
Plug in the minut(-) before the 3. Remember that the car is moving the other way, so it has to stop and start moving in the opposite direction before it can catch up with the other one(velocity wise.)

Try:

6.3=-3+2.9(t)

Find t.

Hubert
 
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