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Chapter 2: Problem 6

A car travels \(22.0 \mathrm{~m} / \mathrm{s}\) north for \(30.0 \mathrm{~min}\) and then reverses direction and travels \(28.0 \mathrm{~m} / \mathrm{s}\) for \(15.0 \mathrm{~min}\). What is the car's total displacement? a) \(1.44 \cdot 10^{4} \mathrm{~m}\) b) \(6.48 \cdot 10^{4} \mathrm{~m}\) c) \(3.96 \cdot 10^{4} \mathrm{~m}\) d) \(9.98 \cdot 10^{4} \mathrm{~m}\)

Short Answer

Expert verified
Answer: a) \(1.44 \cdot 10^{4} \mathrm{~m}\)

Step by step solution

01

Calculate the time durations in seconds

We're given the time durations in minutes. To convert them into seconds, we need to multiply by \(60 \mathrm{~s/min}\). For 30 minutes, \(t_{1}=30.0 \mathrm{~min} \times 60 \mathrm{~s/min} = 1800 \mathrm{~s}\) For 15 minutes, \(t_{2}=15.0 \mathrm{~min} \times 60 \mathrm{~s/min} = 900 \mathrm{~s}\)
02

Calculate the distance traveled in each segment

We use the formula distance = velocity × time for both segments. First segment: \(d_{1}=v_{1} \times t_{1}=22.0 \mathrm{~m/s} \times 1800 \mathrm{~s}=39600 \mathrm{~m}\) Second segment: \(d_{2}=v_{2} \times t_{2}=28.0 \mathrm{~m/s} \times 900 \mathrm{~s}=25200 \mathrm{~m}\)
03

Calculate the total displacement

Since the car reverses direction in the second segment, we need to subtract the distance traveled in the second segment from the distance traveled in the first segment to get the total displacement. Total displacement = \(d_{1} - d_{2}=39600 \mathrm{~m} - 25200 \mathrm{~m} = 14400 \mathrm{~m}\) Comparing our result with the answer choices provided, we find that the car's total displacement is: a) \(1.44 \cdot 10^{4} \mathrm{~m}\)

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Most popular questions from this chapter

A speeding motorcyclist is traveling at a constant speed of \(36.0 \mathrm{~m} / \mathrm{s}\) when he passes a police car parked on the side of the road. The radar, positioned in the police car's rear window, measures the speed of the motorcycle. At the instant the motorcycle passes the police car, the police officer starts to chase the motorcyclist with a constant acceleration of \(4.0 \mathrm{~m} / \mathrm{s}^{2}\) a) How long will it take the police officer to catch the motorcyclist? b) What is the speed of the police car when it catches up to the motorcycle? c) How far will the police car be from its original position?

You ride your bike along a straight line from your house to a store \(1000 .\) m away. On your way back, you stop at a friend's house which is halfway between your house and the store. a) What is your displacement? b) What is the total distance traveled? After talking to your friend, you continue to your house. When you arrive back at your house, c) What is your displacement? d) What is the distance you have traveled?

A girl is riding her bicycle. When she gets to a corner, she stops to get a drink from her water bottle. At that time, a friend passes by her, traveling at a constant speed of \(8.0 \mathrm{~m} / \mathrm{s}\). a) After \(20 \mathrm{~s}\), the girl gets back on her bike and travels with a constant acceleration of \(2.2 \mathrm{~m} / \mathrm{s}^{2} .\) How long does it take for her to catch up with her friend? b) If the girl had been on her bike and rolling along at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\) when her friend passed, what constant acceleration would she need to catch up with her friend in the same amount of time?

An object starts from rest and has an acceleration given by \(a=B t^{2}-\frac{1}{2} C t,\) where \(B=2.0 \mathrm{~m} / \mathrm{s}^{4}\) and \(C=-4.0 \mathrm{~m} / \mathrm{s}^{3}\). a) What is the object's velocity after 5.0 s? b) How far has the object moved after \(t=5.0\) s?

The position versus time for an object is given as \(x=A t^{4}-B t^{3}+C\) a) What is the instantaneous velocity as a function of time? b) What is the instantaneous acceleration as a function of time?
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