Question

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

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

Step-by-step solution

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, \(t_{2}=15.0 \mathrm{~min} \times 60 \mathrm{~s/min} = 900 \mathrm{~s}\)

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: \(d_{2}=v_{2} \times t_{2}=28.0 \mathrm{~m/s} \times 900 \mathrm{~s}=25200 \mathrm{~m}\)

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) \(1.44 \cdot 10^{4} \mathrm{~m}\)

Key concepts

Velocity and Time Relationship

The relationship between velocity and time is crucial when calculating an object's displacement or distance traveled. Velocity is a vector quantity, consisting of both the magnitude (speed) and a direction. It tells us how fast an object is moving and in which direction. In the problem, the car moves at a velocity of 22.0 m/s north initially and then 28.0 m/s in the opposite direction. Knowing the time for each part of the journey allows us to determine how far the car has traveled. To calculate the distance covered in each segment, we multiply the velocity by the time. First, we convert time from minutes to seconds (as seconds are the standard unit of time in physics):

• For the first segment: 30 minutes is converted to 1800 seconds.
• For the second segment: 15 minutes is converted to 900 seconds.

Understanding this relationship helps in calculating displacement, especially when an object travels over different velocities and time intervals.

Direction Reversal in Motion

When analyzing motion, direction plays an important role. A reversal in direction affects how we calculate overall displacement. Displacement is concerned not just with the total distance traveled, but also with the final position relative to the starting point. In our exercise, the car initially travels north, then reverses. This means the second part of the journey moves opposite to the first. This requires us to subtract the distance covered in the return journey from the initial northward travel to determine the final displacement:

• The car travels 39,600 meters north first.
• Then, it travels 25,200 meters in the opposite direction.

Thus, calculating displacement involves subtracting the reversed direction distance from the initial forward distance, ensuring accurate representation of position change.

Distance Versus Displacement

Understanding the difference between distance and displacement is crucial in physics problems like these. Distance refers to the total path traveled by an object, regardless of direction. Displacement, however, measures the net change in position, taking direction into account. In the given problem, the car's total travel adds up as follows:

• Distance: 39,600 meters + 25,200 meters equals 64,800 meters.
• Displacement: Calculated as 39,600 meters north minus 25,200 meters south, resulting in 14,400 meters northward.

Thus, while the car covered a total distance of 64,800 meters, its displacement is only 14,400 meters north. This illustrates how displacement can be significantly less if the object reverses direction during its travel.