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An athlete swims the length \(L\) of a pool in a time \(t_{1}\) and makes the return trip to the starting position in a time \(t_{2} .\) If she is swimming initially in the positive \(x\) direction, determine her average velocities symbolically in (a) the first half of the swim, (b) the second half of the swim, and (c) the round trip. (d) What is her average speed for the round trip?

Short Answer

Expert verified
The average velocities for the first half, second half, and the round trip are \( \frac{L}{t_{1}}\), \(\frac{-L}{t_{2}}\) and 0 respectively. The average speed for the round trip is \(\frac{2L}{t_{1} + t_{2}}\).

Step by step solution

01

Calculate the average velocity during \(t_{1}\)

To calculate the average velocity during \(t_{1}\), we will use the formula for average velocity which is total displacement over total time. Since we are regarding the outward swim as movement along the positive x direction, the displacement is \(L\) and the total time is \(t_{1}\). So, \(v_{avg1} = \frac{L}{t_{1}}\).
02

Calculate the average velocity during \(t_{2}\)

The swimmer covers the same distance \(L\) during \(t_{2}\), but in the opposite direction, therefore, the displacement is \(-L\). Using the same formula for average velocity, the velocity during the second half is \(v_{avg2} = \frac{-L}{t_{2}}\).
03

Calculate the average velocity for the round trip

For the entire round trip, the total displacement is zero, as the swimmer has returned to the starting position. Therefore, even with the formula \(v_{avg} = \frac{total\ displacement}{total\ time}\), the average velocity will be zero as the numerator is zero. So, \(v_{avg} = 0\).
04

Calculate the average speed for the round trip

The formula to calculate average speed is total distance divided by total time. The total distance the swimmer swam is \(2L\) and the total time spent is \((t_{1} + t_{2})\). So, \(speed_{avg} = \frac{2L}{t_{1} + t_{2}}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement and Distance
Understanding the difference between displacement and distance is paramount in comprehending motion dynamics. Displacement refers to a vector quantity that depicts the change in position of an object. It's directed from the starting point to the final position, and possesses both magnitude and direction. In contrast, distance measures the total ground covered, regardless of the direction, and is a scalar quantity.

For instance, if a runner completes a lap around a 400-meter track, their displacement is zero, as they return to the starting point, but the distance covered is 400 meters. In physics problems such as our example with the swimmer traversing a pool back and forth, the displacement for the entire trip is zero (since the swimmer returns to the starting point), yet the distance is the sum of the lengths of each individual swim, doubling the single length of the pool. This distinction is crucial when calculating average speed versus average velocity.
Kinematics in Physics
Kinematics is a branch of mechanics that describes the motion of points, objects, and systems of bodies without considering the forces that cause the motion. It involves studying the trajectories of objects, as well as their velocities and accelerations, which are often symbolized as paths on graphs.

In kinematics, average velocity is a vector quantity that measures the overall rate of change in position, again with direction noted, whereas average speed is a scalar and reflects how fast an object is moving irrespective of its moving direction. In our athlete's scenario, we calculate both the average velocity - which considers direction and results in zero for a round trip where the starting and ending points are the same - and the average speed, which only concerns the total path length traveled over time.
Symbolic Representation in Physics
Physics often relies on symbolic representation to succinctly convey relationships between different elements of a problem. Symbols become a universal language allowing complex equations and concepts to be written in a standardized form. When solving for kinematic quantities like in our swimmer's example, we use symbols such as L for distance, t for time, and v for velocity.

This symbolic method aids in creating general formulas that can apply to various situations. For instance, average velocity is often represented as \( v_{avg} = \frac{displacement}{time} \), whereas average speed is \( speed_{avg} = \frac{distance}{time} \). The symbolic expression offers an efficient way to plug in known values and compute unknowns, making it a fundamental aspect of physics problem-solving and particularly helpful for students to learn and understand these universal patterns in physics equations.

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Most popular questions from this chapter

A model rocket is launched straight upward with an initial speed of \(50.0 \mathrm{~m} / \mathrm{s}\), It accelerates with a constant upward acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) until its engines stop at an altitude of \(150 \mathrm{~m}\). (a) What can you say about the motion of the rocket after its engines stop? (b) What is the maximum height reached by the rocket? (c) How long after liftoff does the rocket reach its maximum height? (d) How long is the rocket in the air?

Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than \(800 \mathrm{~m} / \mathrm{s}^{2}\) lasting for any length of time will not cause injury, whereas an acceleration greater than \(1000 \mathrm{~m} / \mathrm{s}^{2}\) lasting for at least \(1 \mathrm{~ms}\) will cause injury. Suppose a small child rolls off a bed that is \(0.40 \mathrm{~m}\) above the floor. If the floor is hardwood, the child's head is brought to rest in approximately \(2.0 \mathrm{~mm}\). If the floor is carpeted, this stopping distance is increased Io about \(1.0 \mathrm{~cm}\). Calculate the magnitude and duration of the deceleration in both eases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate.

An attacker at the base of a castle wall \(3.65 \mathrm{~m}\) high throws a rock straight up with speed \(7.40 \mathrm{~m} / \mathrm{s}\) at a height of \(1.55 \mathrm{~m}\) above the ground. (a) Will the rock reach the top of the wall? (b) If so, what is the rock's speed at the top? If not, what initial speed must the rock have to reach the top? (c) Find the change in the speed of a rock thrown straight down from the top of the wall at an initial speed of \(7.40 \mathrm{~m} / \mathrm{s}\) and moving between the same two points. (d) Does the change in speed of the downward-moving rock agree with the magnitude of the speed change of the rock moving upward between the same elevations? Explain physically why or why not.

A hockey player is standing on his skates on a frozen pond when an opposing player, moving with a uniform speed of \(12 \mathrm{~m} / \mathrm{s}\), skates by with the puck. After \(3.0 \mathrm{~s}\), the first player makes up his mind to chase his opponent. If he accelerates uniformly at \(4.0 \mathrm{~m} / \mathrm{s}^{2}\), (a) how long does it take him to catch his opponent, and (b) how far has he traveled in that time? (Assume the player with the puck remains in motion at constant speed.)

A truck covers \(40.0 \mathrm{~m}\) in \(8.50 \mathrm{~s}\) while smoothly slowing down to a final velocity of \(2.80 \mathrm{~m} / \mathrm{s}\). (a) Find the truck's original speed. (b) Find its acceleration.

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