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Chapter 2: Problem 10

Two cars travel in the same direction along a straight highway, one at a constant speed of \(55 \mathrm{mi} / \mathrm{h}\) and the other at \(70 \mathrm{mi} / \mathrm{h}\). (a) Assuming they start at the same point, how much sooner does the faster car arrive at a destination \(10 \mathrm{mi}\) away? (b) How far must the faster car travel before it has a 15 -min lead on the slower car?

Short Answer

Expert verified
a) The faster car arrives at the destination 2.34 min earlier than the slower car. b) The faster car has to travel 11.77 mi to get a 15-min lead on the slower car.

Step by step solution

01

Compute Time Taken by Each Car to Reach Destination

To compute the time taken by each car to reach the destination, we use the time formula \( t = \frac{d}{v} \). For the slower car, \( t_{\text{slow}} = \frac{10 \, \text{mi}}{55 \, \text{mi/h}} = 0.182 \, \text{h} \) and for the faster car, \( t_{\text{fast}} = \frac{10 \, \text{mi}}{70 \, \text{mi/h}} = 0.143 \, \text{h}.
02

Compute Time Difference

To find out how much sooner the faster car reaches the destination, we subtract the time taken by the faster car from that taken by the slower car. So, \( t_{\text{diff}} = t_{\text{slow}} - t_{\text{fast}} = 0.182 \, \text{h} - 0.143 \, \text{h} = 0.039 \, \text{h} \). However, since this quantity is usually given in minutes, we convert it by multiplying by 60 to get \( t_{\text{diff}} = 2.34 \, \text{min} \).
03

Calculate the Distance for 15-Min Lead

Using time formula again, we can find the distance the faster car must travel before it has a 15-min (or 0.25 h) lead. This time, we'll equate the time taken by the slower car to reach a certain distance to the time taken by the faster car plus 15 min, i.e., \( t_{\text{slow}} = t_{\text{fast}} + 0.25 \rightarrow d_{\text{slow}} = d_{\text{fast}} = v_{\text{fast}} * (t_{\text{slow}} - 0.25) \). Substituting the values, \( d_{\text{fast}} = 70 \, \text{mi/h} * (0.182 \, \text{h} - 0.25 \, \text{h}) = 11.77 \, \text{mi} \).

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Most popular questions from this chapter

A race car moves such that its position fits the relationship $$ x=(5.0 \mathrm{~m} / \mathrm{s}) t+\left(0.75 \mathrm{~m} / \mathrm{s}^{3}\right) t^{3} $$ where \(x\) is measured in meters and \(t\) in seconds. (a) Plot a graph of the car's position versus time. (b) Determine the instantaneous velocity of the car at \(t=4.0 \mathrm{~s}\), using time intervals of \(0.40 \mathrm{~s}, 0.20 \mathrm{~s}\), and \(0.10 \mathrm{~s}\). (c) Compare the average velocity during the first \(4.0 \mathrm{~s}\) with the results of part (b).

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