# On a 600-mile trip, Andy averaged 5 miles per hour faster for the last 128 miles than he did for the first 472 miles. The entire trip took 10 hours. How fast did he travel for the first 472 miles?

Jun 17, 2018

$\textcolor{b l u e}{64 \text{ mph}}$

#### Explanation:

$\text{speed"="distance"xx"time}$

Let: $t = \text{time", s="speed}$

Andy's average speed for the first part of the journey:

$s = \frac{472}{t} \setminus \setminus \setminus \left[1\right]$

Andy's average speed for the second part of the journey was 5 miles an hour faster:

$s + 5 = \frac{128}{t} \setminus \setminus \setminus \left[2\right]$

Rearranging $\left[1\right] \mathmr{and} \left[2\right]$ in terms of $t$

$t = \frac{472}{s}$

$t = \frac{128}{s + 5}$

Total time is 10 hrs:

$\therefore$

$\frac{472}{s} + \frac{128}{s + 5} = 10$

Solve for $\boldsymbol{s}$

$472 \left(s + 5\right) + 128 s = 10 {s}^{2} + 50 s$

$- 10 {s}^{2} + 550 s + 2360 = 0$

${s}^{2} - 55 s - 236 = 0$

Factor:

$\left(s + 4\right) \left(s - 59\right) = 0 \implies s = - 4 \mathmr{and} s = 59$

Negative speed not applicable here.

$s = 59$

Speed for second part of journey is 5 miles per hr greater:

$59 + 5 = 64 \text{ mph}$

We can check these results.

Time travelling at 59 mph

$\frac{472}{59} = 8 \text{ hrs}$

Time travelling at 64 mph

$\frac{128}{64} = 2 \text{ hrs}$

Total time:

$8 + 2 = 10$