Question

Henry is going on a trip. The first leg of his trip is 32 miles, where he drives at a constant rate. The second leg of his trip is 12 miles, where he drives at half the rate of the first
leg. If the total trip takes 1 hour, how fast does Henry drive during the first leg of his trip?

Answer 1

Henry's rate of movement in the two legs can be determined by applying the formula of speed. Thus, his speed during his first leg is 0.6 hours.

The speed of an object is a measure of how fast it moves in a given time. So that;

Speed = $\frac{Distance}{Time}$

In the given question, let the rate of his fist leg be represented by x. Thus;

⇒ Time = $\frac{Distance}{Speed}$

For the first leg of his trip;

Time taken, $T_{1}$  = $\frac{32}{x}$

For the second leg of his trip;

Time taken, $T_{2}$ = $\frac{12*2}{x}$

                   = $\frac{24}{x}$

Given that the total trip takes 1 hour, then:

$T_{1}$ + $T_{2}$ = 1 hour

$\frac{32}{x}$  + $\frac{24}{x}$ = 1

$\frac{56}{x}$ = 1

x = 56

Then;

$T_{1}$ = $\frac{32}{x}$ = $\frac{32}{56}$

    = $\frac{4}{7}$

$T_{1}$ = $\frac{4}{7}$ hours

$T_{2}$ = $\frac{24}{x}$ = $\frac{24}{56}$

   = $\frac{3}{7}$

$T_{2}$ = $\frac{3}{7}$ hours

Henry's speed during the first leg of his trip is $\frac{4}{7}$ hours (0.6 hours).

Visit: https://brainly.com/question/17140675