Distance Question

Question

You and a friend are racing motorcycles. You take off south going 3 m/s, then 2 seconds later your friend goes west going 4 m/s. Find a formula for D(t), the distance in meters between you and your friend t seconds after you take off. (my hint is that the answer should be a piecewise function)

Answer 1

You are going south and your friend is going west; your two 

travel paths, along with a line representing the distance

between you and your friend make a right triangle

 

Using the pythagorean theorem:

 

(your distance)² + (friend's distance)² = (distance between)²

 

Rate x Time = Distance

 

t is your time

 

t-2  is the time for your friend

 

{(3m/s)(t)}² + {(4m/s)(t - 2)}² = (Distance between)²

 

√[{(3m/s)t }² + {(4m/s)(t - 2)}²]  = Distance between you

and your friend

Answer 2

Hi John

 

Let set the reference of time zero for the friend racing to South. Assuming that he is able to keep a constant velocity all of the time, the distance d1 travelled by him after t seconds is:

 

  d1 = V1*t = 3*t

 

The other friend start racing 2 seconds later. It means that when the first friend who is racing South have a clock indicating t seconds, the second one reads t + 2 seconds. Then, the distance traveled by the second friend is:

 

d2=V2*(t+2) = 4*(t + 2)

 

Because they are racing along two paths forming 90 degrees (South and West), the distance between them is equal to the hypotenuse of a right triangle. It means, if we call D(t) the distance between both friends after t seconds, it is equal to:

 

D(t) = sqrt ( (3t)² + (4(t+2))²) = sqrt ( 9t² + 16t² + 64t + 64)

 

D(t) = sqrt ( 25t² + 64t + 64)

 

I hope it helps you.