Pete and Kathy are standing at home on a softball field. Pete starts walking up the third base line at the rate of 6 feet per second. At the same time, Kathy starts jogging up the first base line at the rate of 10 feet per second. In how many seconds will they be 270 feet apart?

Did you make a diagram?

Let the time taken be t seconds,
so the distance along third base line is 6t feet, and the distance along the first base line is 10t feet.

hoping that the lines at home plate are at 90º, don't you have a right-angled triangle and therefore isn't

(6t)^2 + (10t)^2 = 270^2 ?

t = .....

(pretty big softball field !)

Thanks I was using the triangle, but I was just plugging in numbers trying to find some that worked, I did not think of making an equation. Thank you!

To find the time it takes for Pete and Kathy to be 270 feet apart, we can use the concept of relative velocity.

Let's assume that Pete and Kathy start from the same point, and they move in opposite directions along the third base line and the first base line, respectively. Since they are moving away from each other, we can add their velocities to get the relative velocity.

The relative velocity between Pete and Kathy is the sum of their individual velocities:
Relative velocity = Pete's velocity + Kathy's velocity

Relative velocity = 6 feet per second + 10 feet per second
Relative velocity = 16 feet per second

Now we can use the formula 'distance = rate × time' to find the time it takes for them to be 270 feet apart.

Distance = Relative velocity × Time

Plugging in the values:
270 feet = 16 feet per second × Time

Now solve for Time:
Time = 270 feet / 16 feet per second

Calculating the value:
Time = 16.875 seconds

Therefore, it will take Pete and Kathy approximately 16.875 seconds to be 270 feet apart.