It doesn't make sense to get negative

[davedave]

The speed of car A is 72.2 km/h while the speed of car B is 53 km/h. If car B is now 48 km ahead of car A, how much time is needed for car A to catch up with car B?


car A
let x = the current position of car A
speed = 72.2 km/h
time = x/72.2 hours

car B
let x + 48 = the current position of car B
speed = 53 km/h
time = (x + 48)/53 hours

so, both cars will have traveled for the same amount of time when car A catches up with car B.

then, x/72.2 = (x + 48)/53

solving for x gives x = -180.5 km
thus, time of car A is -2.5 hours.

You cannot have negative time.

Could someone explain my solution? Thanks.

 

[Hurkyl]

What does "time of car A" mean, and what does it have to do with the problem?

 

[CAF123]

Perhaps start with a picture of the problem and take the intial position of A to be at the origin to simplify things a little. Then write the position as a function of time for both cars. If you do this and be careful with signs, that should correct your error.

 

[jfgobin]

Davedave, instead of expressing the "times", write:

$A\left ( t \right ) = ...$ (position of car A at time $t$)
$B\left ( t \right ) = ...$ (position of car B at time $t$)
​

And express what "to catch up" means in terms of these functions.

 

[HallsofIvy]

The speed of car A is 72.2 km/h while the speed of car B is 53 km/h. If car B is now 48 km ahead of car A, how much time is needed for car A to catch up with car B?car A
let x = the current position of car A

There is no point in including this at all. Set up your "coordinate system" so that when t= 0, x= 0.

speed = 72.2 km/h
time = x/72.2 hours

This is time to go "x" km but doesn't tell you anything. "x" is not relevant to the problem.

car B
let x + 48 = the current position of car B

Since we are taking x= 0, this is just 48 km ahead of A.

speed = 53 km/h
time = (x + 48)/53 hours

Again, this is just the time B would have taken to get to its current position and is irrelevant.

so, both cars will have traveled for the same amount of time when car A catches up with car B.

then, x/72.2 = (x + 48)/53

No, these are the times until they reach their current positions (with A 48 km behind B) and you don't know that they are the same because you don't know where they started.

Instead do this in either of two ways:
1) Taking A's current position as our x= 0 point in t hours A will have gone 72.2t km and be at position x= 72.2t. In that same t hours, B will have gone 53t km and will be at x= 53t+ 48. Set those equal and solve for t.

2) Since speed is "relative", do everything "relative to B", treating B as if it were standing still. A is "closing on B" at a relative speed of 72.2- 53= 19.2 km per hour. How long will it take A to cover the 48 km between it and B?

solving for x gives x = -180.5 km
thus, time of car A is -2.5 hours.

You cannot have negative time.

Could someone explain my solution? Thanks.

No, because your "solution" is based on faulty reasoning.