How Do You Calculate When Two Accelerating Cars Meet?

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SUMMARY

The problem involves calculating the meeting point of two cars, a Honda and a Porsche, with differing acceleration rates. The Honda accelerates at 4.00 m/s² and reaches a speed of 23.0 m/s before the Porsche, which accelerates at 8.00 m/s², begins its motion. The Honda travels 66.125 meters during its initial acceleration phase. To determine when the Porsche catches up, one must set up equations based on their respective distances traveled over time, incorporating their acceleration and initial velocities.

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How would you solve a problem such as this?

A Honda and a Porsche race, starting from the same point. The Honda accelerates at a constant 4.00 m/s2; the Porsche at a constant 8.00 m/s2. The Porsche gives the Honda an advantage by letting it start first. The Honda accelerates, and when it is traveling at 23.0 m/s, the Porsche starts. How far do the cars travel from the starting point before the Porsche catches up with the Honda?

All I have is this:

honda v = 4x = 23 m/s
x = 5.75 s

honda position = 2x^2
honda (5.75) = 2(5.75)^2 = 66.125 m

so all I know is when the honda is 66.125 meters the porche starts to go.

what do I next?
 
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Now, you imagine that you are in the Porsche. You see yourself accelerating towards the Honda at 8-4=4 m/s^2 and have to travel 66.125 m before catching up to it.
 
When they meet after time t,
Honda travels a distance
d = vo*t + 0.5*a1*t^2 Here vo = 23 m/s ...(1)
Porsche travels a distance
d + 66.125 = vo*t + 0.5*a2*t^2. Here vo = 0...(2)
From eq.1 and 2, find t.
 

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