Solving the distance formula for rate and time
Solving the distance formula for rate and time is a Grade 11 Algebra 1 skill from enVision Chapter 1 that rearranges d = rt to isolate any variable. To solve for rate: r = d/t. To solve for time: t = d/r. For a car traveling 240 miles in 4 hours: r = 240/4 = 60 mph. For a train at 75 mph covering 180 miles: t = 180/75 = 2.4 hours. For 450 km in 5 hours: r = 90 km/h. This foundational equation-solving skill applies to motion, speed, and travel problems throughout algebra and real-world contexts.
Key Concepts
The distance formula $d = rt$ can be solved for any variable: To solve for rate: $r = \frac{d}{t}$ To solve for time: $t = \frac{d}{r}$.
Common Questions
What is the distance formula?
d = rt, where d is distance, r is rate (speed), and t is time. It relates the three quantities in uniform motion problems.
How do you solve d = rt for rate?
Divide both sides by t: r = d/t. For 240 miles in 4 hours: r = 240/4 = 60 mph.
How do you solve d = rt for time?
Divide both sides by r: t = d/r. For 180 miles at 75 mph: t = 180/75 = 2.4 hours.
A trip of 450 km takes 5 hours. What is the average speed?
r = 450/5 = 90 km/h.
How far does a train travel at 80 mph for 3.5 hours?
d = rt = 80 * 3.5 = 280 miles.
How do you use d = rt when distance and rate are given but time is unknown?
Rearrange to t = d/r, then substitute the known values. Always check units (miles vs km, hours vs minutes) are consistent.