# Joan left Durham traveling 85 mph. Mike,to catch up,left some time later driving at 94 mph. Mike caught up after 7 hours. How long was Joan driving before Mike caught up?

Apr 26, 2018

Joan had been driving for $7$hours $44$mins

#### Explanation:

*Before we start we should immediately note that $\text{speed, distance and time}$ are all present in the question. Therefore we know we will most likely need to use the equation: $\text{Speed"="Distance"/"Time}$

Firstly let's call the place where the two people meet $x$.

To get to point $x$ Joan travelled at $85 m p h$ for $y$ amount of time.

To get to point $x$ Mike travelled at $94 m p h$ for $7$ hours.

Mike has given us both $\text{speed and time}$ We can, therefore, now work out the distance Mike meets Joan.
-Rearrange the formula :$\text{Distance"="Speed" times "Time}$

$\text{Distance} = 94 \times 7$

$\text{Distance"=658"miles}$

We now know that the point they both meet at (which we called $x$ to start with is: $658$

We now know the $\text{Distance and Speed}$ of Joan. We can, therefore, work out the length of time Joan has been travelling.
-Rearrange the formula: $\text{Time"="Distance"/"Speed}$
$\text{Time} = \frac{658}{85}$

$\text{Time"=7.74"hours} \left(2. D . P .\right)$

Now all that remains for us to do is work out how many minutes Joan has been travelling for.

$\frac{\text{mins}}{60} \times 100 = 74$

$\frac{\text{mins}}{60} = \frac{74}{100}$

$\text{mins} = \frac{74}{100} \times 60$

$\text{mins} = 44$

Joan has therefore been travelling for $7$hours $44$mins