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Shkiper50 [21]
3 years ago
11

If a car on cruise control moves at a speed of 50 mi/hr for 3 hours, how far did they travel?

Physics
1 answer:
sergij07 [2.7K]3 years ago
7 0

Answer:

150 miles

Explanation:

Distance = Speed × Time

50 mph × 3 h = 150 miles

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Orbital velocity is the average speed of a planet moving through space in its orbit around the sun. Which of the following plane
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3 years ago
Traveling 221 miles from Boston Back Bay Station to NYC Penn Station takes 3 hours
Nostrana [21]

Answer:

Approximately 116\; \text{miles} for the train from Boston to NYC Penn Station.

Approximately 105\; \text{miles} for the train from NYC Penn Station to Boston.

Explanation:

Convert minutes to hours:

\begin{aligned}t(\text{BOS $\to$ NYC}) &= 3\; {\text{hour}} + 40\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &=\left(3 + \frac{2}{3}\right)\; \text{hour}\\ &= \frac{11}{3}\; \text{hour} \end{aligned}.

\begin{aligned}t(\text{NYC $\to$ BOS}) &= 4\; {\text{hour}} + 5\; \text{minute} \times \frac{1\; {\text{hour}}}{60\; \text{minute}} \\ &= \frac{49}{15}\; \text{hour} \end{aligned}.

Calculate average speed of each train:

\begin{aligned}v(\text{BOS $\to$ NYC}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{11}{3}\; \text{hour}} \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}.

\begin{aligned}v(\text{NYC $\to$ BOS}) &= \frac{s}{t}\\ &= \frac{221\; \text{mile}}{\displaystyle \frac{49}{15}\; \text{hour}} \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}\end{aligned}

Assume that it takes a time period of t for the trains to pass by each other after departure. Distance each train travelled would be:

s(\text{NYC $\to$ BOS}) = v(\text{NYC $\to$ BOS})\, t.

s(\text{BOS $\to$ NYC}) = v(\text{BOS $\to$ NYC})\, t.

Since the trains have just passed by each other, the sum of the two distances should be equal to the distance between the stations:

v(\text{NYC $\to$ BOS})\, t + v(\text{BOS $\to$ NYC})\, t = 221\; \text{mile}.

Rearrange and solve for t:

(v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC}))\, t = 221\; \text{mile}.

\begin{aligned}t &= \frac{221\; \text{mile}}{v(\text{NYC $\to$ BOS}) + v(\text{BOS $\to$ NYC})} \\ &= \frac{221\; \text{mile}}{\displaystyle \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} + \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1}} \\ &= \frac{539}{279}\; \text{hour}\end{aligned}.

Distance each train travelled in t = (539 / 279)\; \text{hour}:

\begin{aligned}s(\text{BOS $\to$ NYC}) &= v\, t \\ &= \frac{663}{11}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 116\; \text{mile}\end{aligned}.

\begin{aligned}s(\text{NYC $\to$ BOS}) &= v\, t \\ &= \frac{2652}{49}\; \text{mile} \cdot \text{hour}^{-1} \times \frac{539}{279}\; \text{hour} \\ &\approx 105\; \text{mile} \end{aligned}.

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