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?

Question

Eva Adamson · Accepted Answer

#color(blue)(64 " mph")#

#"speed"="distance"xx"time"#

Let: #t="time", s="speed"#

Andy's average speed during the initial segment of the trip:

#s=472/t \ \ \[1]#

Andy traveled the second leg of the trip at an average speed that was five miles per hour faster:

#s+5=128/t \ \ \[2]#

Rearranging #[1] and [2]# in terms of #t#

#t=472/s#

#t=128/(s+5)#

It takes ten hours total:

#:.#

#472/s+128/(s+5)=10#

Solve for #bbs#

10s^2+50s=4772(s+5)+128s

#-10s^2+550s+2360=0#

#s^2-55s-236=0#

A factor

#(s+4)(s-59)=0=>s=-4 and s=59#

Here, negative speed is not relevant.

#s=59#

Speed on the second portion of the trip is five miles per hour faster:

#59+5=64 " mph"#

We can verify these outcomes.

Traveling at 59 miles per hour

#472/59=8 " hrs"#

Time at 64 miles per hour

#128/64=2" hrs"#

Time in total:

#8+2=10#

Hazel Anglin · Answer

undefined Let's denote Andy's speed for the first 472 miles as $ x $ miles per hour. Since he averaged 5 miles per hour faster for the last 128 miles, his speed for the last 128 miles would be $ x + 5 $ miles per hour.

The total time taken for the trip is 10 hours, so we can set up the equation:
$$ \frac{472}{x} + \frac{128}{x + 5} = 10 $$

Now, solve for $ x $. After simplifying and solving the equation, we find that $ x = 52 $ miles per hour.

So, Andy traveled at a speed of $ 52 $ miles per hour for the first 472 miles.