Homework Help: Motion Equation Problem?

1. Sep 16, 2009

r_swayze

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?

2. Sep 16, 2009

ideasrule

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.

3. Sep 16, 2009

rl.bhat

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.