A can run 3/4 times as fast as B .If A gives B a start of 60 metres, how far must the winning post be in order that A & B reach it at the same time?
2007-02-04 02:22:18 · 4 answers · asked by abc 1 in Science & Mathematics ➔ Mathematics
Your question got problem.
Lets say B give A a start of 60 meters.
Because B run faster than A.
When B run v, A only run 3/4v
s/v = (s-60)/(3/4v)
3s = 4s-240
s = 240 meters (from B)
2007-02-04 02:33:13 · answer #1 · answered by seah 7 · 2⤊ 0⤋
this is supposing you mean that A is 3/4 times faster then B
draw a line a mark a spot where B would start at 60 and x being the distance B has to go to the finish.
there for..
B would have to go x or B=x
A would have to go 60 and x or A= 60+x
since a goes 3/4 faster then b you can express that by A= 3/4B
by substitution....
using A=60+x and A=3/4B
60 + x = 3/4 B since we know B=x
60 + x = 3/4 x subtract x
60 = -1/4 x multiply each side by 4
240 = -x you cant have a negative distance
x would equal 240
so the finish line is 300 away from the start( where A starts) and 240 away from where B starts.
2007-02-04 10:46:19 · answer #2 · answered by Cassandra H 2 · 1⤊ 0⤋
A will never reach B. If A is 3/4 as fast, then A is slower than B. If B starts first, its not even a race.
2007-02-04 10:33:01 · answer #3 · answered by ? 1 · 0⤊ 0⤋
If A does not run as fast as B, how can A give B a start?
Or am I missing something?!
2007-02-04 10:29:50 · answer #4 · answered by Como 7 · 0⤊ 0⤋