1. ## speed, distance calculations

A traffic helicopter is flying with an average speed of 144 km/h over a traffic jam on
the motorway. It takes 7.5 minutes flying in the direction of traffic and 6.5 minutes in
the opposite direction. What is the length of the traffic jam and the average velocity
at which the cars move?

so relative velocity = distance/60 km/s

518400-velocity of cars =distance/60 km/s

2. Originally Posted by Duke
A traffic helicopter is flying with an average speed of 144 km/h over a traffic jam on
the motorway. It takes 7.5 minutes flying in the direction of traffic and 6.5 minutes in
the opposite direction. What is the length of the traffic jam and the average velocity
at which the cars move?

so relative velocity = distance/60 km/s

518400-velocity of cars =distance/60 km/s
I am unsure how how you got the above answers. You should have a system of equations and be careful about the mixed units.

You should let d be the length of the traffic jam and v be the speed of traffic

$d=\left(\frac{12}{5}-v \right)\left( \frac{15}{2}\right) \quad d=\left(\frac{12}{5}+v \right)\left( \frac{13}{2}\right)$

3. How did you obtain those equations?

4. Originally Posted by Duke
How did you obtain those equations?
distance equals rate times time

$d=rt$

5. If you were trying to get the speed of the heli into km/minute should you not have multiplied by 60. And I think you have the signs muddled

6. Originally Posted by Duke
If you were trying to get the speed of the heli into km/minute should you not have multiplied by 60. And I think you have the signs muddled
ummm no

$144 \frac{\text{km}}{\text{hr}}\cdot \frac{1 \text{hr}}{60 \text{min}}=\frac{12 \text{km}}{5\text{min}}$

Please explain why you think they are wrong and what is your solution?