Álgebra 1Solving EquationsA-CED.A.1

Distance, Rate and Time Problems

Use d = rt with a table to set up motion problems.

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Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.

Motion problems use one formula, $d = rt$: distance equals rate times time. You have used rates like miles per hour when converting units, and you can write equations from words. Here the two skills meet.

Por qué funciona

A rate of $50$ miles per hour means $50$ miles for every hour. In $3$ hours that is $50 + 50 + 50 = 50 \cdot 3$ miles. So distance is rate times time, as long as the units match: if the rate is per hour, the time must be in hours.

Idea clave

Make a table with a row for each traveler and columns for rate, time and distance, filling the distance as rate $\times$ time. Then write one equation from how the distances (or times) are related: they add, they are equal, or they differ.
Two cars driving apart: their two distances together make the total gap.

Worked examples

Ejemplo resuelto: Opposite directions

Two cars leave the same town at the same time, one going east at $50$ mph and one going west at $30$ mph. When are they $160$ miles apart?
  1. Let $t$ be the time in hours. East: rate $50$, distance $50t$. West: rate $30$, distance $30t$.
  2. Moving apart, the distances add: $50t + 30t = 160$.
  3. $80t = 160$, so $t = 2$.
  4. Check: $100 + 60 = 160$ miles.

Respuesta: After $2$ hours.

Ejemplo resuelto: Catching up

A truck leaves at $45$ mph. One hour later a car follows on the same road at $60$ mph. How long does the car take to catch up?
  1. Let $t$ be the car's time in hours. The truck has driven one hour longer: $t + 1$.
  2. Truck: $45(t + 1)$. Car: $60t$. Same direction, so at the moment it catches up the distances are equal: $60t = 45(t + 1)$.
  3. Distribute: $60t = 45t + 45$. Subtract $45t$: $15t = 45$, so $t = 3$.
  4. Check: the car goes $180$ miles; the truck goes $45 \cdot 4 = 180$ miles.

Respuesta: $3$ hours.

Ejemplo resuelto: A round trip

Lena bikes to a lake at $12$ mph and back at $8$ mph. The trip takes $2.5$ hours in all. How far is the lake?
  1. Let $d$ be the one-way distance. Time is distance divided by rate: $\frac{d}{12}$ there and $\frac{d}{8}$ back.
  2. The times add: $\frac{d}{12} + \frac{d}{8} = 2.5$.
  3. Multiply every term by $24$: $2d + 3d = 60$, so $5d = 60$ and $d = 12$.

Respuesta: $12$ miles.

Error común

Adding the speeds when both travelers go the same way. Rates add only when the distances add (moving apart or toward each other). When one catches the other, the distances are equal. A second slip is mixing units: $20$ minutes at $36$ mph is $36 \cdot \frac{1}{3} = 12$ miles, not $36 \cdot 20$.

Práctica

Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 3 XP.

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Este es uno de los peldaños más altos de Álgebra 1.

Fuentes y lecturas

  • OpenStax, Elementary Algebra 2e, section 3.5 Solve Uniform Motion Applications
  • Common Core State Standards for Mathematics, A-CED.A.1