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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