A car travels at 60mph, a truck travels 55mph in the opposite direction. The vehicles pass in .34 seconds. How long is the truck?

How do I set this problem up??

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- Oct 18th 2012, 09:41 AM #1

- Oct 18th 2012, 11:19 AM #2

- Oct 18th 2012, 01:48 PM #3

- Oct 18th 2012, 03:36 PM #4

- Oct 19th 2012, 02:09 AM #5

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## Re: How do I set up this problem?

The solutions above assume that the time it takes for the vehicles to pass is measured from the moment when their front bumpers align to the moment when the front bumper of the car aligns with the back of the truck. During this time, the car travels the whole truck's length relative to the truck. This is why the rate of closure times 0.34 s presumable equals the length of the truck. However, at the end of this time, the car and the truck still overlap. If we measure the passing time as the time when the vehicles overlap, then the rate of closure times 0.34 s equals the combined lengths of the vehicles.

- Oct 19th 2012, 06:39 AM #6

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## Re: How do I set up this problem?

Hello, Ranger SVO!

A car travels at 60mph, a truck travels 55mph in the opposite direction.

The vehicles pass in 0.34 seconds. .How long is the truck?

Is that a misprint?

If this is conventional "passing vehicles", there is no unique solution.

. . We would need the length of the car.

Normally, the passing procedure begins when their front bumpers are even

. . and ends when their*rear bumpers are even.*

Code:* - - * | car | → * - - * Start * - - - - * ← | truck | * - - - - * * - - * | car | → End * - - * * - - - - * ← | truck | * - - - - *

Let $\displaystyle T$ = length of the truck.

They have a combined speed of $\displaystyle 115$ mph.

It is as if the car has*stopped*

. . and the truck is passing at 115 mph.

The truck travels $\displaystyle 5280(C+T)$ feet at $\displaystyle \tfrac{506}{3}$ ft/sec

. . and takes 0.34 seconds.

We have: .$\displaystyle 5280(C+T) \:=\:\tfrac{506}{3}(0.34) \quad\Rightarrow\quad C+T \:=\:0.03258333$ feet.

The combined lengths of both vehicles is*less than a half-inch!*

And we still don't know the value of $\displaystyle T.$

- Oct 19th 2012, 07:06 AM #7

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- Oct 19th 2012, 04:04 PM #8

- Oct 21st 2012, 08:19 AM #9

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