Question

Joe drove at the speed of 45 miles per hour for a certain distance. He then drove at the speed of 55 miles per hour for the same distance. What is the average speed for the whole trip?

Answer 1

Answer:

$v_{avg} = 49.5 mph$

Explanation:

Let the distance moved by Joe is "d"

so the time taken by him to drove it by speed 45 mph is given as

$t_1 = \frac{d}{v_1}$

$t_1 = \frac{d}{45}$

now the same distance is traveled by him with speed 55 mph

so the time taken by him

$t_2 = \frac{d}{55}$

so total time taken by him for complete distance 2d

$t = t_1 + t_2$

$t = \frac{d}{45} + \frac{d}{55}$

$t = 0.0404 d$

now the average speed is given as

$v_{avg} = \frac{2d}{t}$

$v_{avg} = \frac{2d}{0.0404d}$

$v_{avg} = 49.5 mph$

Answer 2

Answer:

47.37 miles per hour,

Explanation:

The average speed is given by the formula

Average speed =(total distance / total time)

Let the distance Joe traveled at a speed of 45 miles per hour be 'D', because he then drove the same distance with a speed of 50 miles per hour then, the total distance is '2D'.

The total time will be the time he drove at a speed of 45 miles per hour ($t_{1}$) plus the time he drove at a speed of 50 miles per hour ($t_{2}$).

Then the average speed is :

$v_{average} = \frac{2D}{t_{1}+t_{2}}$.

Because we know that uniform rectilinear motion is discribed by the ecuation

$d=\v*t$ (where d is distance, v is a constant speed and, t is the time)

we can express both times in terms of speeds and the distance, thus

$t_{1} =\frac{D}{v_{1}}$ and,

$t_{2} =\frac{D}{v_{2}}$.

So now the average speed $v_{average} = \frac{2D}{t_{1}+t_{2}}$ can be written as follows:

$v_{average} = \frac{2D}{(\frac{D}{v_{1}}+\frac{D}{v_{2}} )}$

$v_{average} = \frac{2D}{D(\frac{1}{v_{1}}+\frac{1}{v_{2}} )}$

$v_{average} = \frac{2}{(\frac{1}{v_{1}}+\frac{1}{v_{2}} )}$

$v_{average} = \frac{2}{(\frac{1}{45}+\frac{1}{50} )}$

$v_{average} = 47.37$ miles per hour.