Question

5. A truck left a diner at 1:00 p.M. And drove 360 km to Jersey City. The truck arrived at 2:00 p.M. And drove to Jersey City at an average 7:00 p.M. A car left the same dinner speed of 80 km/h. A. How fast did the truck travel?

Answer 1

360 km/h

Step-by-step explanation:

The truck left the diner at 1.00 pm and arrived at Jersey City at 2.00 pm - This means that the total time of the trip was 1 hour:

`t = 1 h`

The distance covered between the diner and Jersey City was 360 km:

`S=360 km`

So, the average speed of the truck is given by the distance covered divided by the time taken:

`v=(S)/(t)=(360 km)/(1 h)=360 km/h`

Answer 2

average speed of the truck will be 60 km/h

Step-by-step explanation:

Truck started his Journey from diner at 1:00 PM

It reached to the Jersey City at 7:00 PM

so here we have total time interval of journey for the truck motion is given as

`t = 7 - 1 = 6 hour`

total distance = 360 km

so here we know that average speed is defined as

`v_(avg) = (distance)/(time)`

`v_(avg) = (360 km)/(6 h)`

`v_(avg) = 60 km/h`

so average speed of the truck will be 60 km/h