Speed, Distance, and Time

Distance, speed, and time are linked by a single equation, distance equals speed times time, but contest problems rarely hand that equation to you directly. Multi-leg trips hide a tempting shortcut that gives the wrong average speed, and problems about two moving objects require combining their speeds correctly depending on whether they move toward, apart from, or after one another.

What you will learn

• Apply distance equals speed times time
• Handle problems with changing speeds or directions
• Solve meeting and catch-up problems

Keep paper and a pencil nearby. Try each problem before opening a hint, and mark sections reviewed as you work through them.

The distance-speed-time equation

The core relationship is $\text{distance}=\text{speed}\times\text{time}$, and it rearranges to solve for whichever quantity is missing: $\text{time}=\frac{\text{distance}}{\text{speed}}$ or $\text{speed}=\frac{\text{distance}}{\text{time}}$. Before using this equation, check that the units match — a speed given in miles per hour should be multiplied by a time in hours, not minutes, or the result will be off by a factor of 60.

A constant speed makes this equation apply directly to the whole trip. Most contest problems, however, involve more than one leg of travel or more than one moving object, which is where the equation must be applied piece by piece rather than all at once.

Average speed over multiple legs

Average speed for an entire trip is always $\frac{\text{total distance}}{\text{total time}}$ — it is not the simple average of the individual speeds unless each leg takes exactly the same amount of time. A car that travels 60 miles at 30 mph and then another 60 miles at 60 mph spends 2 hours on the first leg and only 1 hour on the second, for a total of 120 miles in 3 hours, giving an average speed of $40$ mph — not the simple average of 30 and 60, which would incorrectly give 45.

The reason the simple average fails is that more time is spent traveling at the slower speed whenever the distances (rather than the times) are equal, which pulls the true average speed below the midpoint of the two speeds. Always compute the time for each leg separately, add the distances and the times separately, and only then divide.

Meeting and catch-up problems

When two objects travel toward each other starting some distance apart, the gap between them closes at the sum of their two speeds, since both objects are contributing to closing the same distance. Two trains 300 miles apart, closing at 40 mph and 60 mph, close the gap at a combined $40+60=100$ mph, meeting after $300\div100=3$ hours.

When one object chases another moving in the same direction, the gap between them changes at the difference of their speeds, since the faster one only gains ground at the rate by which it outpaces the slower one. If the chasing object also starts later or further behind, that starting gap is treated exactly like the distance in a normal distance-speed-time problem, using the difference of the two speeds as the effective closing speed.

See the reasoning unfold

Pause after each step and predict what comes next.

Common mistakes to catch

• Averaging two speeds directly instead of computing total distance divided by total time.
• Adding two speeds for a catch-up problem instead of subtracting them, or the reverse for a meeting problem.
• Mixing time units, such as multiplying a speed in miles per hour by a time given in minutes without converting first.

Try with support

Guided practice

Choose a strategy, take a first step, and use the hints when you need direction.

Put it into practice

Independent practice

Work on your own first. If you reveal a solution, work through it and mark it reviewed.

Bring it together

• Distance equals speed times time; rearrange this equation to solve for time or speed, always matching units first.
• Average speed for a multi-leg trip is total distance divided by total time, not the simple average of the individual speeds.
• Two objects moving toward each other close their gap at the sum of their speeds; a chaser gains on a target moving the same direction at the difference of their speeds.