Question

Jackie wants to meet up with her friend who lives 3 miles away and bring him the cookies she baked. she and her friend start moving toward each other at the same time. Jackie's walking speed is 3mph, and her friend's biking speed is 12mph. 4 minutes after Jackie left the house, her mom notices that she forgot the cookies on the kitchen table. She sends Jackie's brother to bring them to Jackie. How fast should Jackie's brother travel to catch up with Jackie before she meets her friend?

Answer 1

First, find the time when Jackie and her friend meet.

Using:

`\sf s=ut+\frac12at^2`

where:

• s = displacement
• u = initial velocity
• a = acceleration
• t = time (in hours)

Jackie and her friend are not accelerating, so a = 0, which means the formula can be reduced to:

`\sf s=ut`

Create equations

Jackie:
`\sf s_1= 3t`

Friend:
`\sf s_2 = 12t`

As the distance covered by Jackie and her friend at the time they meet equals 3 miles:

`\sf s_1+s_2=3 \ miles`

`\sf \implies 3t + 12t = 3`

`\sf \implies 15t = 3`

`\sf \implies t = 0.2 \ hr`

`\sf \implies t=0.2 * 60=12 \ mins`

Therefore, they meet 12 mins after they both start moving.

3 mph = 3 ÷ 60 = 0.05 miles per min

12 mph = 12 ÷ 60 = 0.2 miles per min

Therefore,

Jackie traveled: 0.05 x 12 = 0.6 miles

Her friend traveled: 0.2 x 12 = 2.4 miles

when they meet.

To find the speed Jackie's brother must walk to catch up with her at the same time she meets her friend, use s = 0.6 miles (as this is the distance Jackie walked) and t = 12 - 4 = 8 mins (since he left 4 mins after Jackie left):

`\sf s=ut`

`\sf \implies u = (s)/(t)`

`\implies \sf u = (0.6)/(8)`

`\implies \sf u = 0.075 \ miles \ per \ min`

`\sf \implies u = 0.075 * 60 = 4.5 \ mph`

So Jackie's brother needs to walk faster than 4.5 mph to catch up with Jackie before she meets her friend.

(If he walks at 4.5 mph, he will catch up to Jackie at the instant she meets her friend. If he walks slower than 4.5 mph, he will not catch up with Jackie before she meets her friend).