mike can run the mile in 6 minutes and dan can run the mile in 9 minutes. If mike gives dan a head start of 1 minute, how far from the start will mike pass dan?

mike speed = (1/6) mile/min

dan speed = (1/9) mile/min
t-1 = mike time
t = dan time
distance = rate * time = same for both
d = (1/6) (t-1) = (1/9)(t)
3 t -3 = 2 t
t = 3 minute
distance = 3/9 = 1/3 mile

ratemike=1/6 ratedan=1/9

distancemike=ratemike*time
distancedan=ratedan(time-1)
set these equal, and solve for time.
Then, use the first equation to find out how far mike traveled. time will be in minutes, distance in miles.

To determine when Mike will pass Dan, we can calculate their relative speeds. Given that Mike can run the mile in 6 minutes and Dan can run it in 9 minutes, we can find their rates of running by taking the reciprocal of their respective times.

Mike's speed: 1 mile / 6 minutes = 1/6 mile per minute
Dan's speed: 1 mile / 9 minutes = 1/9 mile per minute

Since Mike gives Dan a head start of 1 minute, we know that Dan will already be ahead by a distance equal to his speed multiplied by the time, which is (1/9) mile/minute * 1 minute = 1/9 mile.

Now, to determine how long it will take for Mike to cover this distance and catch up to Dan, we can create an equation:

Mike's distance = Dan's distance + head start distance

Let's assume this time is t minutes:

(1/6)t = (1/9)t + 1/9

Now, we can solve this equation to find the time when Mike passes Dan:

Multiplying by the common denominator of 18 to simplify the equation:

3t = 2t + 2

Combining like terms:

3t - 2t = 2

Simplifying:

t = 2

Therefore, Mike will pass Dan 2 minutes after they both start running. To find how far from the start Mike passes Dan, we can substitute this value of t into the equation:

Distance = Mike's speed * t

Distance = (1/6) * 2

Distance = 1/3 mile

Therefore, Mike will pass Dan 1/3 mile from the starting point.

To find out how far from the start Mike will pass Dan, we first need to determine the distance both individuals can cover in the time given.

Since Mike can run a mile in 6 minutes, his running speed is 1 mile / 6 minutes = 1/6 mile per minute.

Similarly, since Dan can run a mile in 9 minutes, his running speed is 1 mile / 9 minutes = 1/9 mile per minute.

Since Dan gets a head start of 1 minute, he would have covered a distance of 1/9 mile after that minute.

Now, let's assume that after t minutes, Mike catches up with Dan. In that amount of time, Mike would have covered a distance of (1/6) * t, and Dan would have covered a distance of (1/9) * (t - 1) (since he had a 1-minute head start).

To find out when Mike catches up with Dan, we can set these two distances equal to each other and solve for t:

(1/6) * t = (1/9) * (t - 1)

Multiplying both sides of the equation by 18 (the least common multiple of 6 and 9) to eliminate the fractions, we get:

3t = 2(t - 1)

Simplifying,

3t = 2t - 2

Subtracting 2t from both sides,

t = -2

Since we can't have a negative time, this means that Mike never catches up with Dan, and they will never be at the same position along the track. Hence, Mike will never pass Dan.