Question

NEED HELP!

Graph in comments

Kathleen and Arnob both run from the park entrance along a loop. Kathleen starts walking from the park entrance and gets a 5-mile head start on Arnob. The graph shows how far they have both traveled.

How many minutes does it take Arnob to catch up to Kathleen?

a. 9.8
b. 10
c. 73.5
d. 75

Answer 1

ok what grade r u in

Answer 2

10th

Answer 3

oh sorry i cant help i am in 6th grade sorry

Answer 4

:(

Answer 5

well that's easy enough: just looking at the graph, K and A are at the same distance (therefore caught up to each other) at around 73 ~ 75 ish minutes, yeah?
where the red and blue lines intersect)

Answer 6

so to work out if the answer is C or D:
speed = distance over time
so look at the red line
A travels 8 miles in 60 minutes, so his speed is 8 miles per hour
K travels 2 miles in 30 minutes, so her speed is 4 miles per hour
K has a 5 mile head-start, so when does their distance match?

Answer 7

A's distance = initial distance from gate + (speed x time)
A's distance = 0 + (8 x time)

K's distance = initial distance from gate + (speed x time)
K's distance = 5 + (4 x time)

when they're the same distance from the gate, they've caught up with one another
so when:
A's distance = K's distance
0 + (8 x time) = 5 + (4 x time)
8t = 5 + 4t
8t - 4t = 5
4t = 5
t = 5/4 hours = 1 hr and 15 mins = 75 mins

Answer 8

one more way of looking at it:
equation of blue line (K) is \(\large y = \frac1{15}x + 5\)
equation of red line (A) is \(\large y = \frac2{15}x\)
so when red line = blue line, they're caught up
so when:
\(\Large \color{blue} {\frac1{15}x + 5} = \color{red} {\frac2{15}x}\)
\(\Large \color{blue} { 5} = \color{red} {\frac2{15}x}- \color{blue} {\frac1{15}x} \)
\(\Large { 5} = \frac1{15}x \)
\(\Large 5 \times 15 = \frac1{15}x \times 15 \)
\(\Large { 75} = x \)
hope this helps?