Question

Two straight roads diverge at an angle of 45°. two cars leave the intersection at 2:00 p.m., one traveling at 36 mi/h and the other at 52 mi/h. how far apart are the cars at 2:30 p.m.?

Answer 1

Let's take the x-axis as the direction of car 1, which is traveling at speed $v_1 = 36 mi/h$. Car 2 is traveling at speed $v_2 = 52 mi/h$ in a direction of $\alpha = 45 ^{\circ}$ with respect to the x-axis.

Let's start by finding the distance covered by the two cars in between 2.00 pm and 2.30 pm, so in 30 minutes. We can write 30 minutes as 0.5 hours, so $t=0.5 h$. The distance covered by the two cars is
$S_1 = v_1 t = (36 mi/h)(0.5 h)= 18 mi$
$S_2 = v_2 t =(52 mi/h)(0.5 h)=26 mi$
 S1 lies on the x-axis, while we have to decompose S2 on both axes:
$S_{2x}=S_2 \cos \alpha = (26 mi)(\cos 45^{\circ})=18.4 mi$
$S_{2y}=S_2 \sin \alpha = (26 mi)(\sin 45^{\circ})=18.4 mi$

Assuming the two cars started their motion from the origin of the axes (0,0), then we can rewrite the final coordinates of the two cars as
- car 1: (18 mi,0)
- car 2: (18.4 mi, 18.4 mi)
And so the distance between the two cars will be
$d= \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}= \sqrt{(18.4-18)^2+(18.4-0)^2}=$
$=18.4 mi$