Question

You drive 6.0 km at 50 km/h and then another 6.0 km at 90 km/h. Your average speed over the 12 km drive will be _________

Answer 1

64.2857 km/h

Explanation:

Time taken to drive the first 6.0 km in hours = 6/50

Time take to drive another 6.0 km in hours = 6/90

Total time taken to drive 12 km in hours = (6/50) + (6/90)

= 6(50+90)/50x90

= 6x140/50x90

= 840/4500

The average speed over the 12 km is therefore 12 divided by 840/4500

= (12 x4500)/840

= 64.2857 km/h

Answer 2

`64.29\ (km)/(h)`

Explanation:

hello

the speed is the relation between a distance traveled and the time used for it and it is given by the equation

`v =(displacement)/(time)\\ (1)v=(d)/(t)`

Now, the average speed will be

`v_(average) =(displacement\ total)/(total\ time)\\ (1)v=(d)/(t) \\(2)v_(average)=(d_(t) )/(t_(t) )`

Step 1

I drive 6.0 km at 50 km/h

from (1) it is possible to isolate t

`(1)v=(d)/(t)\\t=(d)/(v)`

Let

`v=50 (km)/(h) \\d=6.0 km\\t=(d)/(v) \\t=(6.0 km)/(50 (km)/(h))\\t_(1)=0.12\ hours\\`

Step 2

then I drive 6.0 km at 90 km/h

`v=90 (km)/(h) \\d=6.0 km\\t=(d)/(v) \\t=(6.0 km)/(90 (km)/(h))\\t_(2)=0.067\ hours\\`

Step 3

FInd total time and total displacement

`Total\ time=t_(1) +t_(2) \\t_(t) =0.12\ hours +0.067\ hours =0.187\ hours\\\\total\ distance\ =d_(1) +d_(2)=6.0\ km +6.0\ km= 12\ km\\`

step 4

finally, put these values into the equation(2)

`v_(average)=(d_(t) )/(t_(t) )\\v_(average)=(12\ km )/(0.187\ h )\\v_(average)=64.29\ (km)/(h)`

V=64.29 km/h

Have a good day