Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B , moving in the same straight line at a constant speed of 15 meters per second. how far must car A travel from this initial position before it catches up with car B?
200m, 400m, 800m, 1000m

Solution:

Car A, moving in a straight line at a constant speed of 20 m/sec.

Car A is initially 200 meters behind car B.

Car B is moving in the same straight line at a constant speed of 15 m/sec.

First, let's express the movement of Car A and B in terms of their position over time (relative to car B)

For car A: y = 20x - 200

Car A moves 20 meters every second x, and starts 200 meters behind car B.

For Car B: y = 15x

Car B moves 15 meters every second x and starts at our base point

Now, set both the equations equal to one another to find the time x at which they meet:

15x = 20x - 200

Let us solve for x by transposing. We get

20x - 15x = 200

5x = 200

x = 200/5

x = 40

At time x = 40 seconds, the cars meet.

So, car A traveled at this time,

Car A moves 20 meters every second,

= 20 x 40

= 800 meters

Therefore, car A must travel 800 meters from this initial position before it catches up with car B.

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Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B , moving in the same straight line at a constant speed of 15 meters per second. how far must car A travel from this initial position before it catches up with car B?

Summary:

Car A, moving in a straight line at a constant speed of 20. meters per second, is initially 200 meters behind car B , moving in the same straight line at a constant speed of 15 meters per second. Car A must travel 800 meters from this initial position before it catches up with car B.