SOLUTION: respected sir, 1] A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10 th of a minute. In the second heat, A gives B a head start

Algebra ->  Customizable Word Problem Solvers  -> Travel -> SOLUTION: respected sir, 1] A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10 th of a minute. In the second heat, A gives B a head start      Log On

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Question 198257: respected sir,
1] A and B ran a race of 480 m. In the first heat, A gives B a head start of 48 m and beats him by 1/10 th of a minute. In the second heat, A gives B a head start of 144 m and is beaten by 1/30 th of a minute. What is B’s speed in m/s?
plz solve my doubt..
thank u.
Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
What the problem doesn't say, and must be assumed is that
A and B each travel at constant speeds. I can then write a
constant speed equation for each and for each race:
d%5Ba%5D+=+r%5Ba%5D%2At%5Ba%5D
d%5Bb%5D+=+r%5Bb%5D%2At%5Bb%5D
If a have a stopwatch, I want to start timing the race when
both runners are in motion, so I'll start it when A begins
running for each race.
1st race:
d%5Ba%5D+=+480
d%5Bb%5D+=+480+-+48
d%5Bb%5D+=+432
t%5Ba%5D+=+t%5Bb%5D+-+1%2F10
t%5Bb%5D+=+t%5Ba%5D+%2B+1%2F10
-------------------
480+=+r%5Ba%5D%2At%5Ba%5D (same for both races)
432+=+r%5Bb%5D%2A%28t%5Ba%5D+%2B+1%2F10%29
--------------------
2nd race:
d%5Ba%5D+=+480
d%5Bb%5D+=+480+-+144
d%5Bb%5D+=+336
t%5Bb%5D+=+t%5Ba%5D+-+1%2F30
----------------------
480+=+r%5Ba%5D%2At%5Ba%5D
336+=+r%5Bb%5D%2A%28t%5Ba%5D+-+1%2F30%29
----------------------
r[a] and r[b] are the same in both races.
Also, t[a] is the same in both races, but
t[b] is not the same .
In race 1:
432+=+r%5Bb%5D%2A%28t%5Ba%5D+%2B+1%2F10%29
r%5Bb%5D+=+432%2F%28t%5Ba%5D+%2B+1%2F10%29
I can plug this into the equation from race 2
336+=+r%5Bb%5D%2A%28t%5Ba%5D+-+1%2F30%29
336+=+%28432%2F%28t%5Ba%5D+%2B+1%2F10%29%29%2A%28t%5Ba%5D+-+1%2F30%29
Multiply both sides by t%5Ba%5D+%2B+1%2F10
336%2A%28t%5Ba%5D+%2B+1%2F10%29+=+432%2A%28t%5Ba%5D+-+1%2F30%29
336t%5Ba%5D+%2B+33.6+=+432t%5Ba%5D+-+14.4
96t%5Ba%5D+=+48
t%5Ba%5D+=+.5
The time for A is 30 sec
480+=+r%5Ba%5D%2At%5Ba%5D
480+=+r%5Ba%5D%2A.5
r%5Ba%5D+=+960 m/min
432+=+r%5Bb%5D%2A%28t%5Ba%5D+%2B+1%2F10%29 (1st race)
432+=+r%5Bb%5D%2A%28.5+%2B+.1%29
r%5Bb%5D+=+432%2F.6
r%5Bb%5D+=+720 m/min
720 m/min * 1%2F60 min/sec = 12 m/sec
B's speed is 12 m/sec
---------------------
check answer:
336+=+r%5Bb%5D%2A%28t%5Ba%5D+-+1%2F30%29 (2nd race)
336+=+r%5Bb%5D%2A%2815%2F30+-+1%2F30%29
336+=+r%5Bb%5D%2A%2814%2F30%29
r%5Bb%5D+=+10080%2F14
r%5Bb%5D+=+720 m/min
720 m/min = 12 m/sec
OK
To see the whole thing, I could plot time (sec) on
the x-axis and distance (m) on the y-axis for
I want to look at d+=+480m
d+=+16t (both races)
d+-+48+=+12t B, 1st race
d+-+144+=+12t B, 2nd race
+graph%28+700%2C+400%2C+-10%2C+60%2C+-50%2C+700%2C+16x%2C12x+%2B+48%2C12x+%2B+144%29+
The line starting at the origin is A's plot.
Where d+=+480 shows the times to get to finish
+graph%28+700%2C+400%2C+-10%2C+60%2C+300%2C+700%2C+16x%2C12x+%2B+48%2C12x+%2B+144%29+
This is a blow-up