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NkNewaj
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Hi.. there is small calculation mistake...

Should be \(= \frac{1800}{120} = 15\) m/s[/quote]

Thanks, sashiim20.
I calculated that just before sleeping, perhaps the reason why an error occurred :)
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2 cars race around a circular track, in opposite directions, at constant speeds. They start at the same point and meet every 30.0 seconds. If they move in the
same direction, they meet every 120.0 seconds. If the track is 1800 meters long, what is the speed of first car?

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Speed = Distance / Time

Distance \(= 1800\) m

Lets speed of cars be \(x\) and \(y\)

When cars race in opposite direction, relative speed will be added \(= x + y\)

Time when cars race in opposite direction \(= 30\) s

When cars race in same direction, relative speed will be subtracted \(= x - y\)

Time when cars race in same direction \(= 120\) s

\(x + y = \frac{1800}{30} = 60\) m/s

\(x - y = \frac{1800}{120} = 15\) m/s

\(x + y = 60\) ------- (\(i\))

\(x - y = 15\) -------- (\(ii\))

Adding equations (\(i\)) and (\(ii\)) we get;

\(2x = 75\)

\(x = 37.5\) m/s

Substituting value of \(x\) in (\(i\)), we get;

\(y = 60 - 37.5 = 22.5\) m/s

If first car is the fastest of the two cars, speed of \(x= 37.5\) m/s
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sashiim20

Posted from my mobile device

Hi.. there is small calculation mistake...

Should be \(= \frac{1800}{120} = 15\) m/s

Thanks, sashiim20.
I calculated that just before sleeping, perhaps the reason why an error occurred :)[/quote]

:thumbup:

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