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

Answer:

64.2857 km/h

Step-by-step 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

Answer:

$64.29\ \frac{km}{h}$

Step-by-step explanation:

hello

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

$v =\frac{displacement}{time}\\ (1)v=\frac{d}{t}$

Now, the average speed will be

$v_{average} =\frac{displacement\ total}{total\ time}\\ (1)v=\frac{d}{t} \\(2)v_{average}=\frac{d_{t} }{t_{t} }$

Step 1

I drive 6.0 km at 50 km/h

from (1) it is possible to isolate t

$(1)v=\frac{d}{t}\\t=\frac{d}{v}$

Let

$v=50 \frac{km}{h} \\d=6.0 km\\t=\frac{d}{v} \\t=\frac{6.0 km}{50 \frac{km}{h}}\\t_{1}=0.12\ hours$

Step 2

then I drive 6.0 km at 90 km/h

$v=90 \frac{km}{h} \\d=6.0 km\\t=\frac{d}{v} \\t=\frac{6.0 km}{90 \frac{km}{h}}\\t_{2}=0.067\ hours$

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)

$v_{average}=\frac{d_{t} }{t_{t} }\\v_{average}=\frac{12\ km }{0.187\ h }\\v_{average}=64.29\ \frac{km}{h}$

V=64.29 km/h

Have a good day