ACT Math: Solve Distance-Rate-Time Problems with the D=RT Triangle

The formula Distance=Rate×Time (D=RT) is the foundation for all motion problems. To remember all three forms, draw a triangle with D at the top and R and T at the bottom. Cover the variable you want to solve for; the remaining two multiply (if side-by-side) or divide (if one is above the other). If solving for D: D=R×T. If solving for R: R=D/T. If solving for T: T=D/R. This triangle eliminates the need to memorize three formulas; one diagram gives you all three algebraically.

Example: A car travels at 60 mph for 3 hours. How far did it go? D=60×3=180 miles. Another: A plane traveled 300 miles in 2 hours. What was its speed? R=300/2=150 mph. Another: A runner covered 10 miles at 5 mph. How long did it take? T=10/5=2 hours. Each uses the same triangle rearranged. Drawing the triangle on your test paper takes 5 seconds and ensures you use the correct formula.

Type 1: Simple problems ask for one of D, R, or T given the other two. Use the triangle directly. Type 2: Two-object problems (catching up, meeting, separate). Set D1=D2 (when they meet) or use D1=D2+difference (when one is ahead). Type 3: Average speed problems. Average speed=Total distance/Total time (not the average of the two speeds). Identifying the problem type before you set up equations prevents common errors like adding two speeds instead of using the average formula.

Example Type 2: "Car A leaves at 50 mph. Car B leaves 1 hour later at 60 mph. When do they meet?" At meeting time, distances are equal. Car A's time=t, Car B's time=t-1 (started 1 hour later). Set D_A=D_B: 50t=60(t-1), so 50t=60t-60, so -10t=-60, so t=6 hours. Car B catches up after 6 hours. Example Type 3: A person drives 100 miles at 50 mph, then 100 miles at 100 mph. Average speed=200/time. Time=(100/50)+(100/100)=2+1=3 hours. Average=200/3≈67 mph, not (50+100)/2=75 mph.

Problem 1: A jogger runs 8 miles at 4 mph. How long? Problem 2: A cyclist travels 2 hours at 15 mph. Distance? Problem 3: A train covers 240 miles in 4 hours. Speed? Problem 4: Two cars leave the same point; one at 40 mph, one at 60 mph, in opposite directions. How far apart after 3 hours? Problem 5: A car travels 150 miles at 50 mph, then 100 miles at 100 mph. Average speed? Problem 6: Runner A leaves at 8 mph. Runner B leaves 15 minutes later at 10 mph. When does B catch A? For each, (1) draw the triangle or set up the equation, (2) solve, (3) check your answer. Use the triangle for Problems 1-3; set up equations for Problems 4-6.

Answers: 1) T=8/4=2 hours. 2) D=15×2=30 miles. 3) R=240/4=60 mph. 4) After 3 hours: Car 1 at 120 miles one direction, Car 2 at 180 miles opposite direction. Total separation=300 miles (opposite) or 60 miles (same direction). 5) Total distance=250 miles, total time=2.5+1=3.5 hours. Average≈71.4 mph. 6) At catch-up, distances equal. A's time=t, B's time=t-0.25. 8t=10(t-0.25), so 8t=10t-2.5, so t=1.25 hours.

Distance-rate-time questions appear on nearly every ACT Math test and reward systematic setup. Many students lose points by confusing average speed with the average of two speeds, or by forgetting to account for delayed starts. Mastering the triangle and the three problem types adds 1-2 reliable points per test and builds confidence because DRT problems follow predictable patterns.

Drill the six problems twice this week. By test day, you will draw the triangle and solve DRT problems faster than most students can read them.