Steve McSpoke left home on his bicycle at 8am, traveling at 18km/h. At 10am, Steve's brother set out after him on a motorcycle, following the same route. The motorcycle traveled at 54km/h. How long had steve traveled when his brother overtook him?
Steve's brother time to catch-up --- t hrs
Steve's time to catch-up by brother -- t+2
54t = 18(t+2)
54t = 18t + 36
36t = 36
t = 1
Steve had biked for 3 hours, when his brother overtook him.
thanks
Also, how far had steve traveled when his brother overtook him?
simply sub in the value of t = 1 into both distances., you will obviously get the same answer since they both went the same distance.
Reiny, how can you show me how to do the last part wirten out?
Thank you so much!
neverminds. I got it.
To find out how long Steve had traveled when his brother overtook him, we can set up a basic equation. Let's denote the time Steve traveled as "t" hours.
Since Steve McSpoke left home at 8 am and his brother set out at 10 am, there is a time difference of 2 hours. Therefore, Steve traveled for t hours while his brother traveled for t - 2 hours.
We can use the formula: Distance = Speed × Time
For Steve:
Distance = Speed × Time
Distance = 18 km/h × t
For Steve's brother:
Distance = Speed × Time
Distance = 54 km/h × (t - 2)
Since they both traveled the same distance when his brother overtook him, we can set up an equation:
18t = 54(t - 2)
Now, let's solve for t:
18t = 54t - 108
-36t = -108
t = 3
Therefore, Steve had traveled for 3 hours when his brother overtook him.