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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 jet plane lands with a speed of \(100 \mathrm{~m} / \mathrm{s}\) and can accelerate at a maximum rate of \(-5.00 \mathrm{~m} / \mathrm{s}^{2}\) as it comes to rest. (a) From the instant the plane touches the runway, what is the minimum time needed before it can come to rest? (b) Can this plane land on a small tropi\(\mathrm{cal}\) island airport where the runway is \(0.800 \mathrm{~km}\) long?

A truck tractor pulls two trailers, one behind the other, at a constant speed of \(100 \mathrm{~km} / \mathrm{h}\). It takes \(0.600 \mathrm{~s}\) for the big rig to completely pass onto a bridge \(400 \mathrm{~m}\) long. For what duration of time is all or part of the trucktrailer combination on the bridge?

A speedboat increases its speed uniformly from \(v_{i}=\) \(20.0 \mathrm{~m} / \mathrm{s}\) to \(v_{f}=30.0 \mathrm{~m} / \mathrm{s}\) in a distance of \(2.00 \times 10^{2} \mathrm{~m}\). (a) Draw a coordinate system for this situation and label the relevant quantities, including vectors. (b) For the given information, what single equation is most appropriate for finding the acceleration? (c) Solve the equation selected in part (b) symbolically for the boat's acceleration in terms of \(v_{f}, v_{i}\), and \(\Delta x\). (d) Substitute given values, obtaining that acceleration. (e) Find the time it takes the boat to travel the given distance.

A car starts from rest and travels for \(t_{1}\) seconds with a uniform acceleration \(a_{1} .\) The driver then applies the brakes, causing a uniform acceleration \(a_{2}\). If the brakes are applied for \(t_{2}\) seconds, (a) how fast is the cas going just before the beginning of the braking period: (b) How far does the car go before the driver begins to brake? (c) Using the answers to parts (a) and (b) as the initial velocity and position for the motion of the ca during braking, what total distance does the car travel Answers are in terms of the variables \(a_{1}, a_{2}, t_{1}\), and \(t_{2}\).

A person takes a trip, driving with a constant speed of \(89.5 \mathrm{~km} / \mathrm{h}\), except for a \(22.0\)-min rest stop. If the person's average speed is \(77.8 \mathrm{~km} / \mathrm{h}\), (a) how much time is spent on the trip and (b) how far does the person travel?
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