Question

You drive on Interstate 10 from San Antonio to Houston, half the time at 54 km/h and the other half at 118 km/h. On the way back you travel half the distance at 54 km/h and the other half at 118 km/h. What is your average speed?

Answer 1

Answer:

Average speed: 86 km/h

Explanation:

Driving from San Antonio to Houston:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

Driving way back:

1st. half time: 54km/h

2nd. half time: 118 km/h

Average speed = [tex] \frac{54 \frac{km}{h}+ 118 \frac{km}{h}  }{2}=86 \frac{km}{h} [\tex]

As in both routes we have the same average speed, then the average speed for the whole trip is 86 km/h

Answer 2

Answer:

$v_{avg} = 79.7 km/h$

Explanation:

As we move from San Antonio to Houston

let the distance is "d" from Antonio to Houston

Half the time it moves with 54 km/h and next half the time it moves 118 km/h

so we will have

$54 T + 118 T = d$

$T = \frac{d}{172}$

so total time is

$2T = \frac{d}{86} = 0.0116 d$

now while his return journey half the distance he move with 54 km/h and next half distance with speed 118 km/h

so we have time of return journey

$T' = \frac{d/2}{54} + \frac{d/2}{118}$

so now we have

$T' = 0.0135 d$

now for the average speed we know that

$v_{avg} = \frac{distance}{time}$

$v_{avg} = \frac{d + d}{0.0116 d + 0.0135 d}$

$v_{avg} = 79.7 km/h$