Constant Acceleration of Two Cars

[ff4930]

[SOLVED] Constant Acceleration

Homework Statement

Two cars are traveling along a straight road. Car A maintains a constant speed of 80 km/h and car B maintains a constant speed of 110 km/hr. At t=0, car B is 45 km behind Car A.
A) How much farther will car A travel before Car b overtakes it?
B)How much ahead of A will B be in 30 seconds after it overtakes A?

Homework Equations

The Attempt at a Solution

I am stumped. I read and reread the section on my textbook about constant acceleration and the material is so confusing. Can someone please hint me or point me to a direction of understanding this please. Thanks.

 

[Littlepig]

Why constant acceleration? Acceleration is 0 as you only have constant speed right?

 

[Doc Al]

Come up with expressions for the position of each car as a function of time. I'd call the location of Car b at t = 0 to be the origin. (So Car A is at 45 km at t = 0.)

There's no acceleration mentioned, so this should be easy.

 

[ff4930]

110x = 80x+45
x=.5?
is this the right approach? I am really lost.

 

[belliott4488]

110x = 80x+45
x=.5?
is this the right approach? I am really lost.

Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.

 

[usman27]

At the time instant where two cars meet up, the distance traveled by Car B will be 45 more than Car A...
v = x/t is the main equation for this type of question.

 

[ff4930]

Your equation is right, but how did you get x = .5? And what are the units? For that matter, what is x? What have you solved for here?

If you explain your reasoning a little more, we can help you better.

.5 hr? it will take .5 hr for Car B to catch up to Car A?

So Car A will travel another 40 km before Car B catches up.

 

[belliott4488]

Okay, so x is time (it's more conventional to use "t" for a time interval and "x" for distance, but it's not necessary). I think you just need to be a little more careful in solving your equation for x - you just made an algebra mistake.

 

[ff4930]

Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?

 

[belliott4488]

Oops, thanks, Suppose to be 45/30 = 1.5.
So Car A will travel another 120 km before Car B catches up?

B)asks the distance Car B will be ahead of for 30 seconds after it passes Car A.
Car B 110 km/h(110 km per 3600 seconds), Car A 80 km/hr(80 km per 3600 seconds)
each second for Car B covers .031km and each second for Car A covers .022km
.031km * 30sec = .93 km.

So after 30 seconds, Car B will be .93 km ahead of Car A

Is this approach correct?

Not quite - .031 km/sec * 30 sec gives the distance B has traveled, but A isn't just standing still - it's covered some distance, too. The difference between the two tells you how far B is ahead of A.

 

[ff4930]

Cool! Thanks belliott, much obliged.