Story Problem - Distance, Speed, and Time

[Mikayla2010]

A woman driving a car 14ft long is passing a truck 30ft long. The tuck is traveling at 50mph. How fast must the woman drive her car so that she cna pass the truck completely in 6 seconds?

Just need help with the equation because I don't even know where to start..

 

[Deleted member 4993]

A woman driving a car 14ft long is passing a truck 30ft long. The tuck is traveling at 50mph. How fast must the woman drive her car so that she cna pass the truck completely in 6 seconds?

Just need help with the equation because I don't even know where to start..

I am assuming these are all constant speed problem - no acceleration involved. Is that correct?

Do you understand - relative velocity?

 

[Mikayla2010]

Yes. The speed is constant

 

[Denis]

HINT: as the car starts overtaking the truck, the BACK of the car is 44 feet from the FRONT of the truck.

Now think a bit...

 

[galactus]

Convert to feet per second.

50 mph is \(\displaystyle \frac{50 \;\ \frac{mi}{hr}\cdot 5280 \;\ \frac{ft}{mi}}{3600 \;\ \frac{s}{hr}}=\frac{220}{3} \;\ ft/s\)

In 6 seconds the truck travels \(\displaystyle \frac{220}{3}\cdot 6=440 \;\ ft\)

Therefore, the woman must travel 440+30+14 =484 feet to overtake the truck.

Now, use d=rt to find her rate. Her rate is then \(\displaystyle \frac{484}{6}=\frac{242}{3} \;\ ft/sec\)

In miles per hour, this is \(\displaystyle \frac{242}{3}\cdot \frac{3600}{5280}=\boxed{55 \;\ mph}\)

 

[soroban]

Hello, Mikayla2010!

Here's a back-door approach . . .

A woman driving a car 14 ft long is passing a truck 30 ft long. .The truck is traveling at 50 mph.
How fast must the woman drive her car so that she can pass the truck completely in 6 seconds?

In the diagram, the vehicles are moving to the right.

For the moment, imagine that the truck has stopped.

                  A - - - - - 44-ft - - - - - - - B
      *-----------*                   *-----------*
      |   14 ft   |                   |   14 ft   |
      *-----------*-------------------*-----------*
                  |       30 ft       |
                  *-------------------*

The passing begins when her front bumper is even with the truck's rear bumper ... at \(\displaystyle A.\)

The passing ends when her rear bumper is even with the truck's front bumper.
. . Her front bumper is at \(\displaystyle B.\)

So she must travel 44 feet in 6 seconds. .Her speed must be 44 ft/6 sec.

. . \(\displaystyle \frac{\text{44 ft}}{\text{6 sec}} \times \frac{\text{60 sec}}{\text{1 min}} \times \frac{\text{60 min}}{\text{1 hr}} \times \frac{\text{1 mile}}{\text{5280 ft}} \;=\; \text{5 mph}\)

Hence, she must travels 5 mph faster than the truck.


Therefore, her speed must be: .\(\displaystyle 50 + 5 \,=\,55\text{ mph.}\)


Edit: Corrected a silly typo . . . *blush*

 

[Denis]

Edit: Corrected a silly typo . . . *blush*

Go stand in the corner with Subhotosh...