Lynn and Tina are planning a foot race. Lynn can run 16.9 feet per second and Tina can run 10 feet per second. Lynn gives Tina 50 foot head start.

Question

When will Lynn catch up to Tina and how many feet away from the start will they be?

Answer 1

The first step is to create your linear equations in the form of y = mx+b. If we were graphing, then the y-axis would be for distance and the x-axis would be for time in seconds, so we will use those same variables below (x for time in seconds and y for distance in feet). Remember that "b" is the y-intercept, so the units for any value of b will be distance in feet as well.

For Lynn, her equation is y= 16.9x + 0, as in she moves 16.9 feet per second (the time in seconds is signified here by the variable 'x') plus an added distance of zero feet, as she started with no head start.

Tina's linear equation is y= 10x + 50, as in she moved 10 feet per second (the time in seconds is signified here by the variable 'x) plus an added distance of 50 feet, as Tina was given a head start of 50 feet, so that will always be a constant addend to her speed.

You now need to set the equations equal to each other since we are trying to find at what point their distances will be the same, and solve for x.

16.9x = 10x + 50 set equations equal to each other

-10x -10x combine like terms

6.9x = 0 +50 divide to solve for x

6.9 6.9

x = 7.25 seconds

now, plug 7.25 back into the original linear equations to get your answers.

For Lynn,

y = 16.9x

= (16.9 feet per second)(7.25 seconds)

y = 122.525 feet

For Tina,

y = 10x + 50 feet

= (10 feet per second)(7.25 seconds) + 50 feet

= 72.5 + 50

y = 122.5 feet

Lynn and Tina meet up at 7.25 seconds and at 122.5 feet away from the startline.