Answer:
First, find the time when Jackie and her friend meet.
Using:
[tex]\sf s=ut+\dfrac12at^2[/tex]
where:
Jackie and her friend are not accelerating, so a = 0, which means the formula can be reduced to:
[tex]\sf s=ut[/tex]
Create equations
Jackie: [tex]\sf s_1= 3t[/tex]
Friend: [tex]\sf s_2 = 12t[/tex]
As the distance covered by Jackie and her friend at the time they meet equals 3 miles:
[tex]\sf s_1+s_2=3 \ miles[/tex]
[tex]\sf \implies 3t + 12t = 3[/tex]
[tex]\sf \implies 15t = 3[/tex]
[tex]\sf \implies t = 0.2 \ hr[/tex]
[tex]\sf \implies t=0.2 \times 60=12 \ mins[/tex]
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):
[tex]\sf s=ut[/tex]
[tex]\sf \implies u = \dfrac{s}{t}[/tex]
[tex]\implies \sf u = \dfrac{0.6}{8}[/tex]
[tex]\implies \sf u = 0.075 \ miles \ per \ min[/tex]
[tex]\sf \implies u = 0.075 \times 60 = 4.5 \ mph[/tex]
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).
