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
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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
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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) (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:
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:
------------------- (same for both races)
--------------------
2nd race:
----------------------
----------------------
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:
I can plug this into the equation from race 2
Multiply both sides by
The time for A is 30 sec m/min (1st race) m/min m/min * min/sec = m/sec
B's speed is 12 m/sec
---------------------
check answer: (2nd race) m/min m/min = 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 m (both races) B, 1st race B, 2nd race
The line starting at the origin is A's plot.
Where shows the times to get to finish
This is a blow-up
