# Passing car problem.

• October 15th 2009, 02:15 PM
Alice96
Passing car problem.
I think I found the answers to the first two parts of this question but if you can check it and help me with the third part I would really appreciate it.

Question:
You are driving when a car passes you going 10 fps faster than you. You want to increase your speed to match his.

a)How long will it take, given that you can accelerate at a max of 2fps every second? My answer: 10/2=5 seconds to match speed.

b)If you begin you acceleration at exactly the moment he passes you, how far behind will you be when you match his speed? My answer: 10+8+6+4+2=30 Feet.

c)If you want to be exactly 45 feet behind him at the moment you match his speed, how long do you have to wait after he passes to begin acceleration. My answer: 1.5 seconds given that you will be 30 feet behind if you begin acceleration immediately so waiting 1.5 seconds means he will be an extra 15 feet ahead when you begin acceleration and 30+15 = 45 feet.

All help and input is appreciated!
• October 15th 2009, 02:51 PM
Scott H
I found a different answer for (b).

To find the solution, we frame all our statements in mathematical language. Suppose that we measure distance from the location at which your car begins accelerating, using $y$ to denote your car's distance and $h$ to denote the other car's distance. Based on the information given, two equations hold:

\begin{aligned}
y''&=2\\
h'&=10.
\end{aligned}

I'm not sure if you've learned integration yet, but we can show from these facts that

\begin{aligned}
y&=t^2+vt\\
h&=(10+v)t+s,
\end{aligned}

where $v$ is your initial velocity at the moment you begin accelerating and $s$ is the initial distance of the other car. You are correct that it takes $5$ seconds to accelerate to the speed of the other car. Now, we ask: what is $h(5)-y(5)$ if $s=0$?

Similarly, what must $s$ be in order that $h(5)-y(5)=45$? How long would it take for the other car to make it that far past you?
• October 16th 2009, 02:16 PM
Alice96
Hmmm would the answer to (b) be 20 feet? We haven't learned integration yet so I'm not sure how you got those equations and how would ou use them if the problem doesn't say what the initial velocity is?.
• October 16th 2009, 03:02 PM
Scott H
As it turns out, the initial velocity doesn't matter, since only relative distance matters. (If it helps, think how much our planet's motion would contribute to initial velocity!)

The equations come from the formula:

$s=\frac{1}{2}at^2+v_0t+s_0,$

where $s$ is the distance of a uniformly accelerating object, $v_0$ is its initial velocity, and $s_0$ is its initial distance. This function is the one that has the second derivative

$\frac{d^2s}{dt^2}=a,$

in accordance with the definition of acceleration as the rate of change of the speed of the object.

From the second pair of equations, we see that

\begin{aligned}
h-y&=(10+v)t+s-(t^2+vt)\\
&=10t+vt+s-t^2-vt\\
&=10t-t^2\;\;\;\;\;\;\;\;\;\;\mbox{when }s=0.
\end{aligned}

This function represents the distance between both cars if you begin accelerating at the moment the other car passes you. What is its value at $t=5$?
• October 16th 2009, 03:10 PM
hjortur
Other way to see this is recalling that the formula for placement is:
$
x=x_0 + v_0t + \frac{1}{2}at^2 \quad (1)
$

Now you can think of the problem like this:
You (A) are sitting in a stopped car. Now someone (B) passes you with velocity of 10fps.
You start to accelerate , from 0fps to 10fps over 5 seconds. How far has the other driver
gone after 5 sec, and how far have you gone after 5 secs?

Your starting velocity is 0 and you start at x=0. Plugging that in (1) you get

$
x_A = 0 + 0t + \frac{1}{2}2\cdot5^2= 25
$

Now for $x_B$ you know that he starts at x=0, his starting velocity is 10 and his acceleration is 0. What is $x_B$?
So how far are A and B appart?
• October 16th 2009, 03:16 PM
Alice96
Ahh so at t=5 it would be h - y = 10(5) - 5^2 which is 25. So the car is 25 feet behind it.
That means that if you need to be 45 feet behind it you would have to wait to start accelerating for 2 seconds so that it can get an additional 20 feet ahead.

I hope that this is correct (Worried)
• October 16th 2009, 06:17 PM
Scott H
You got it! :)