Question

On a car trip you drive for 2 hours and 41 minutes on a highway at a speed of 107.0 km/h. Then you stop at a gas station to fill up your tank. You also eat a quick lunch. The whole break lasts 23 minutes. After the break you start your engine up and you switch to a state road. You drive for another 3 hours and 31 minutes at a speed of 67.0 km/h before you arrive to your destination. What was your average speed for the whole trip with the lunchbreak included

Answer 1

v = 79.3 km/h

Step-by-step explanation:

• By definition, the average speed, is the quotient between the total distance traveled and the time needed to travel that distance.
• The total time, is the sum of three times: one while driving before stopping at the gas station (t₁), the time spent there (t₂) and the time since leaving the gas station until reaching the final destination (t₃) .
• Let's convert these times to seconds first:

`t_(1) = 161 min* (60s)/(1min) = 9660 s (1)`

`t_(2) = 23 min* (60s)/(1min) = 1380 s (2)`

`t_(3) = 211 min* (60s)/(1min) = 12660 s (3)`

`t_(tot) =t_(1) +t_(2) +t_(3) = 9660s + 1380s + 12660s = 23700s (4)`

• In order to find the total distance traveled, we need to add the distance traveled before stopping at the gas station (x₁) and the distance traveled after leaving it (x₂).
• Applying the definition of average speed, we can find these distances as follows:

`x_(1) = v_(1) * t_(1) (5)`

`x_(2) = v_(2) * t_(3) (6)`

• where v₁ = 107.0 km/h, and v₂= 67.0 km/h
• As we did with time, let's convert v₁ and v₂ to m/s:

`v_(1) = 107.0 km/h*(1000m)/(1km)*(1h)/(3600s) = 29.7 m/s (7)`

`v_(2) = 67.0 km/h*(1000m)/(1km)*(1h)/(3600s) = 18.6 m/s (8)`

• Replacing (7) and (1) in (5) we get x₁, as follows (in meters):

`x_(1) = v_(1) * t_(1) = 29.7 m/s * 9660 s = 286902 m (9)`

• Doing the same for x₂ with (3) and (8):

`x_(2) = v_(2) * t_(3) = 18.6 m/s * 12660 s = 235476 m (10)`

• Total distance traveled is just the sum of (9) and (10):

`x_(tot) = x_(1) +x_(2) = 286902 m + 235476 m = 522378 m (11)`

• As we have already said, the average speed is just the quotient between (11) and (4), as follows:

`v_(avg) =(\Delta x)/(\Delta t) = (522378m)/(23700s) = 22.0 m/s (12)`

• Converted back to km/h:

`v_(avg) = 22.0 m/s*(1km)/(1000m)*(3600s)/(1h) = 79.3 km/h (13)`