Speed Required to Pass a Truck in 3 Seconds

[bergausstein]

please help set up the equation for this problem,

a car 15 ft.long overtakes a truck 30 ft. long which is traveling at the rate of 45mph. How fast must the car travel to pass the tuck in 3 seconds ?

 

[MarkFL]

How far must the car travel relative to the truck?

 

[bergausstein]

the effective speed of the car is 45-x is this correct?

 

[MarkFL]

the effective speed of the car is 45-x is this correct?

Shouldn't the car be moving faster than the truck?

If you consider my original question, I am trying to guide you to the answer. :D

The car must move a certain distance relative to the truck in 3 seconds. Once you know this distance, you can plug into the relation between distance constant speed and time, to get the speed of the car relative to the truck. Just be mindful of your units, and make sure they match up.

 

[bergausstein]

the car must travel 30 ft. relative to the truck?

 

[MarkFL]

the car must travel 30 ft. relative to the truck?

The car begins the passing maneuver with its front even with the back of the truck.

If the car moves 30 ft. relative to the truck then the front of the car is located where in relation to the truck?

 

[bergausstein]

at the back of the truck.

 

[MarkFL]

at the back of the truck.

No, that's where the front of the car is at the beginning of the passing maneuver. If it moves 30 ft. relative to the truck, then the front of the car is even with the front of the truck (because the truck is 30 ft. long), but we want the back of the car to be even with the front of the truck. So, how far must the car move relative to the truck to complete the passing maneuver?

 

[bergausstein]

the car should travel a total distance of 45 ft.

 

[MarkFL]

the car should travel a total distance of 45 ft.

Correct! (Star)

And we know it must do so in 3 seconds. So, what speed (in mph) is required to move 45 ft. in 3 seconds? Once you find this speed, then add it to that of the truck.

 

[bergausstein]

I'm more comfortable using ft/sec

15ft/sec+66ft/sec = 81ft/sec --->>

but I still have question. why do we have to add it to the speed of the truck?

 

[MarkFL]

Because that 15 ft/sec is in addition to the speed of the truck. The car has to be moving 15 ft/sec faster than the truck to pass it in 3 seconds. :D

 

[bergausstein]

Because that 15 ft/sec is in addition to the speed of the truck. The car has to be moving 15 ft/sec faster than the truck to pass it in 3 seconds. :D

:)(Dance)

 

[soroban]

Hello, bergausstein!

A car 15 ft. long overtakes a truck 30 ft. long which is
traveling at the rate of 45 mph. How fast must the car
travel to pass the truck in 3 seconds?

At the start, the front of the car
is even with the back of the truck.

Start
     *--------*
     |   15   | →
     *--------*
              *--------------*
              |      30      | →
              *--------------*

Three seconds later, the back of the car
is even with the front of the truck.

Finish
                             *--------*
              : - -  30  - - |   15   | →
                             *--------*
              *--------------*
              |      30      | →
              *--------------*

Relative to the truck, the car has moved 45 feet.

Let [tex]x[/tex] = speed of the car (ft/sec).

The truck's speed is: .[tex]45\text{ mph} \,=\,66\text{ ft/sec}[/tex]

Relative to the truck, the car's speed is [tex](x - 66)[/tex] ft/sec.

It is as if the truck is stopped
and the car is passing at [tex](x-66)[/tex] ft/sec.

The car moved 45 feet in 3 seconds at [tex](x-66)[/tex] ft/sec.

. . [tex]3(x-66) \:=\:45 \quad\Rightarrow\quad x \:=\:81[/tex]The speed of the car is [tex]81[/tex] ft/sec

. . or: .[tex]81\cdot\tfrac{60}{88} \,=\,\tfrac{1215}{22} \,=\,55\!\tfrac{5}{22}\text{ mph.}[/tex]