how many laps... math problem

[eddy2020]

hi, dear friends. i struggle with math word problems. i need a lot of practice and i have set out to learn how to sspot patterns and how to solve them. i will need your total help for this. i am thanking anyone in advance for any help.

this is the problem i'm working on right now.
If a is travelling at 120 mph and is one lap (3/4 mile) behind, and B is travelling 110 mph, how many laps will it take A to catch up?

thanks a lot. if you can drop a hint here and there i will try to do my best.
thanksss!

[Baltuilhe]

Good morning!

Space equation to both:
A ==> [tex]S_A=120t[/tex]
B ==> [tex]S_B=110t+0.75[/tex]

So, when S_A=S_B:
[tex]120t=110t+0.75\\
120t-110t=0.75\\
10t=0.75\\
t=0.075[/tex]

Now, we have the 'position':
S_A=120(0.075)=9miles

If one lap has 3/4=0.75 miles:
[tex]laps=\frac{9}{0.75}=12[/tex]

[Guest]

A is traveling 10 mph faster than B so it will take (3/4 mi)/(10 mph)= 3/40 hour to catch up to B. At 120 mph, A will have traveled (3/40)(120)= 9 miles Since each lap is 3/4 mi long, that is 9/(3/4)= 9(4/3)= 12 laps.