Question

If a cyclist increases his speed by 3 miles per hour, he can reduce the time required for a 6 mile trip by 10 minutes. What is the slower speed?

Answer 1

Answer:

Slower speed = 9 mph

Step-by-step explanation:

Let the initial speed (slower speed) = s mph

Speed 's' = $\frac{\text{Distance}}{\text{Time}}$

            s = $\frac{6}{t}$ -------(1)

If the cyclist increases the speed by 3 mph then the final speed = (s + 3) mph

He can reduce the time 't' by 10 minutes ($\frac{1}{6}$ hours)

Now, (s + 3) = $\frac{6}{t-\frac{1}{6}}$

s + 3 = $\frac{36}{6t-1}$ --------(2)

Substitute the value of 's' from equation (1) to the equation (2).

$\frac{6}{t}+3=\frac{36}{6t-1}$

$\frac{2}{t}+1=\frac{12}{6t-1}$

$\frac{2}{t}-\frac{12}{6t-1}=-1$

$\frac{2(6t-1)-12t}{t(6t-1)}=-1$

$\frac{12t-2-12t}{t(6t-1)}=-1$

$-\frac{2}{t(6t-1)}=-1$

6t² - t = 2

6t² - t - 2 = 0

6t² - 4t + 3t - 2 = 0

2t(3t - 2) + 1(3t - 2) = 0

(2t + 1)(3t - 2) = 0

t = $-\frac{1}{2},\frac{2}{3}$ hours

But the time can't be negative.

Therefore, t = $\frac{2}{3}$ hours is the answer.

From equation (1),

s = $\frac{6}{\frac{2}{3}}$

s = 9 miles per hour.

Slower speed = 9 mph is the answer.