The distance covered by the hare and the tortoise in t seconds are 8t and 5t respectively. (Simple Speed-Distance-Time relation)
The tortoise gets a 510m headstart so at t=0 is 1490m.
The functions representing the distance of both of them from the finish line is,
F(x)=2000-8t,. for hare
G(x)=1490-5t,. for tortoise
From the given ordered pair representing Jacinda's progress, her speed is equal to 10 miles per hour. Letting x be the time it takes for Jacinda to catch up with Raj, the total distance Raj would travel is
1.5 + 4.5x
and Jacinda's total distance is 10x. Mathematically,
1.5 + 4.5x = 10x
The value of x from the equation is 0.27 hours. Therefore, Jacinda needs to run 2.7 miles to catch up with Raj.
(-2.8)+0+(-2.6)+6.4
-2.8-2.6+6.4
-5.4+6.4
1
$1500 - $65.06 = $1434.94
Hope I helped!
~ Zoe
