Jorden leaves home for school on his bike everyday at 6:00 AM, traveling at 32 km/h. one day he forgot his homework, which his mom discovered in the car 15 mins after he had left. She immediatey chased after Jorden, traveling at 52 km/h. At what time did she catch up with Jorden.

We solve this question making linear functions for the distances that Jorden and his mom have driven, and then we equalize them to find the moment they caught up.

Jorden leaves home for school on his bike everyday at 6:00 AM, traveling at 32 km/h.

Her mom will have left 15 minutes after he left, when he will already have traveled [tex]\frac{15*32}{60} = \frac{32}{4} = 8\text{km}[/tex]

So, his position after t minutes can be modeled by the following function:

[tex]J(t) = 8 + 32t[/tex]

His mom

His mom goes after him starting from home, so the initial position is 0, with a velocity of 52 km/h. So her position can be modeled by the following function:

[tex]M(t) = 52t[/tex]

At what time did she catch up with Jorden,

She catches up t minutes after 6:15 AM.

t is found when

[tex]M(t) = J(t)[/tex]

So

[tex]8 + 32t = 52t[/tex]

[tex]20t = 8[/tex]

[tex]t = \frac{8}{20}[/tex]

[tex]t = 0.4[/tex]

0.4 of an hour is 0.4*60 = 24 minutes.

So she catched up with Jordan after 6:15 AM + 24 minutes = 6:39 AM