Answer
They pass each other on the road at 4:00 p.m.
Work Step by Step
1. Convert the speeds into miles per minute:
$$45 \space mi/h \times \frac{60 \space min}{1 \space h} = \frac{45}{60} \space mi/min = \frac{3}{4} \space mi/min$$
$$40 \space mi/h \times \frac{60 \space min}{1 \space h} = \frac{40}{60} \space mi/min = \frac{2}{3} \space mi/min$$
2. Determine the equations.
y = distance from Kingstown
x = time (in minutes) counting from 2:15 p.m.
Anthony's position:
Since he starts 15 minutes earlier, we will use $(x + 15)$ in the equation.
$$y = \frac 34 (x + 15) = \frac 34 x+ \frac {45} 4$$
Helen's position:
Since she starts at Queensville, her initial position is 160 mi, and since she is going from Queensville to Kingstown, her speed is negative, because the distance from Kingstown is decreasing.
$$y = 160 - \frac 23 x$$
3. Find the value of "x", so they are in the same position:
$$\frac 34 x + \frac {45} 4 = 160 - \frac 23 x$$
- Multiplying both sides by 12:
$$9 x + 135 = 1920 - 8x$$
$$9x + 8x = 1920 - 135$$
$$17x = 1785$$
$$x= \frac{1785}{17} = 105 $$
Therefore, they will pass each other 105 minutes after 2:15 p.m.
$105 \space minutes = 1 \space h \space and \space 45 \space min$
2:15 p.m. + 1h and 45 min = 4:00 p.m.
