# This is not my homework I am just curious

• High_Voltage
In summary, the conversation discusses a problem where two cars start from rest and accelerate uniformly, with one car starting 6 seconds later. The goal is to determine when the two cars reach the same distance. The conversation provides a kinematic equation and shows how to set up and solve the problem using this equation. The final answer is calculated to be 26 seconds.
High_Voltage

## Homework Statement

A car starts from rest and accelerates uniformly at 3.0 m/s^2 a second car starts from rest 6 seconds later at the same point and accelerates uniformly at 5 m/s^2. How long does it take the second car to overtake the first car.

## Homework Equations

d=vit+(1/2)at^2 I know that there might be a quadratic equation to solve for x.

## The Attempt at a Solution

I came across this question at this website: http://answers.yahoo.com/question/i...6cAp8NojzKIX;_ylv=3?qid=20090917181717AAiO4kQ and I have been stuck on it for about 20 minutes.

Again this is not my homework, but I do need help.

Think about it. The two cars are basically going through the same process. Each starts from rest and begins accelerating. The same equation will apply to both cars. The only difference is when each car starts. Think about how you can write the kinematic equation for the second car relative to the time of the first car.

This is what I have so far 1.5t^2=2.5t^2-30t+90 is this right?

So you want to know when their distances are equal

∆s=vit+.5at^2

∆s1=1.5t^2
∆s2=2.5(t-6)^2 because the second driver is 6 seconds behind

then just set the two to be equal

t^2-30t+90=0

looks like you got it, nice work!

t=26 by my calculation

Mindscrape said:
So you want to know when their distances are equal

∆s=vit+.5at^2

∆s1=1.5t^2
∆s2=2.5(t-6)^2 because the second driver is 6 seconds behind

then just set the two to be equal

t^2-30t+90=0

looks like you got it, nice work!

t=26 by my calculation
Thank you, when you say ∆s=vit+.5at^2 do you mean d=vit+(1/2)at^2?

I suppose you could call it d. The kinematic equation is really

s=si+vi5+.5at^2

where s is the ending point and si is the starting point. Of course, for this case we can define the start at 0, and both have the same start. However, if the other car was 10m ahead then we would have to use the kinematic equation I listed.

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