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Chapter 2: Problem 28

(a) What is the magnitude of the average acceleration of a skier who, starting from reaches a speed of 8.0 \(\mathrm{m} / \mathrm{s}\) when going down a slope for 5.0 \(\mathrm{s} ? \quad\) (b) How far does the skier travel in this time?

Short Answer

Expert verified
(a) 1.6 \( \mathrm{m/s^2} \); (b) 20 m.

Step by step solution

01

Identify Given Information

We are provided with the initial velocity \( u = 0 \, \mathrm{m/s} \), final velocity \( v = 8.0 \, \mathrm{m/s} \), and time \( t = 5.0 \, \mathrm{s} \). Our goal is to find the average acceleration and the distance traveled.
02

Calculate Average Acceleration

To find the average acceleration \( a \), use the formula \( a = \frac{v-u}{t} \). Substitute \( v = 8.0 \, \mathrm{m/s} \), \( u = 0 \, \mathrm{m/s} \), and \( t = 5.0 \, \mathrm{s} \).\[a = \frac{8.0 \, \mathrm{m/s} - 0 \, \mathrm{m/s}}{5.0 \, \mathrm{s}} = \frac{8.0}{5.0} \, \mathrm{m/s^2} = 1.6 \, \mathrm{m/s^2}\]
03

Determine Distance Traveled

To find the distance \( s \), use the equation \( s = ut + \frac{1}{2}at^2 \). Substitute \( u = 0 \, \mathrm{m/s} \), \( a = 1.6 \, \mathrm{m/s^2} \), and \( t = 5.0 \, \mathrm{s} \).\[s = 0 \, \mathrm{m/s} \times 5.0 \, \mathrm{s} + \frac{1}{2} \times 1.6 \, \mathrm{m/s^2} \times (5.0 \, \mathrm{s})^2 = \frac{1}{2} \times 1.6 \times 25 \, \mathrm{m} = 20 \, \mathrm{m}\]

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Key Concepts

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

Average Acceleration
Average acceleration is a measure of how quickly velocity changes over time. In other words, it's the rate at which the speed of an object is increasing or decreasing.
This concept is crucial for understanding how fast a skier can pick up speed when going downhill.
The formula to calculate average acceleration (\[a\]) is straightforward: \[a = \frac{v-u}{t}\] where \(v\) is the final velocity, \(u\) is the initial velocity, and \(t\) is the time taken for this change.
  • The given example involves a skier where the initial velocity \(u\) is \(0 \, \mathrm{m/s}\) because the skier starts from rest.
  • The final velocity \(v\) is \(8.0 \, \mathrm{m/s}\).
  • The time \(t\) during which the acceleration occurs is \(5.0 \, \mathrm{s}\).
By plugging these values into the formula, the skier's average acceleration comes out to be \(1.6 \, \mathrm{m/s^2}\). This tells us that the skier's speed increases by \(1.6 \, \mathrm{m/s}\) every second.
Velocity
Velocity is the speed of an object in a certain direction.
For skiers, velocity is important because it tells not just how fast they are moving but also in which direction鈥攖ypically downhill.
Unlike speed, velocity is a vector quantity which means it has both magnitude and direction.
  • Initial velocity \(u\) is what the skier starts with. In this example, it's \(0 \, \mathrm{m/s}\) because he starts from rest.
  • Final velocity \(v\) indicates how fast the skier is moving after a period of time. Here, it's \(8.0 \, \mathrm{m/s}\).
Understanding velocity helps us grasp the overall movement of the skier. In kinematics, calculating changes in velocity using acceleration and time gives us deeper insights into a moving object's behavior.
Distance Traveled
Distance traveled refers to how far an object has moved along its path. For a skier descending a slope, determining the distance helps in understanding the journey from start to end.
In physics, especially in kinematics, we often calculate this using the following formula:\[s = ut + \frac{1}{2}at^2\]where \(s\) is the distance, \(u\) is the initial velocity, \(t\) is the time, and \(a\) is the average acceleration.
  • In our skier's case, \(u = 0\) because the skier is starting from rest.
  • Substituting \(a = 1.6 \, \mathrm{m/s^2}\) and \(t = 5.0 \, \mathrm{s}\) yields a distance \(s = 20 \, \mathrm{meters}\).
This calculation shows that the skier travels \(20\) meters downhill in \(5\) seconds. Understanding this helps in planning or observing the ski path and its length.
Time
Time, in the context of kinematics, is the duration over which motion occurs. In our skier example, time is how long it takes them to get from a state of rest to moving at a certain speed, and to travel a certain distance down a slope.
Time is a crucial factor because it works hand in hand with velocity and acceleration to determine the motion of an object.
  • Given as \(t = 5.0 \, \mathrm{s}\) in this example, it's the time span over which the skier experiences acceleration, and covers distance.
  • It's used in calculations to find both average acceleration and distance traveled, showing its fundamental role in motion equations.
Without knowing the time, we wouldn't be able to accurately calculate how fast an object is accelerating or how far it has traveled. It's a key component in the equations that describe how objects move in physics.

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Most popular questions from this chapter

A football player, starting from rest at the line of scrimmage, accelerates along a straight line for a time of 1.5 s. Then, during a negligible amount of time, he changes the magnitude of his acceleration to a value of 1.1 \(\mathrm{m} / \mathrm{s}^{2}\) . With this acceleration, he continues in the same direction for another 1.2 \(\mathrm{s}\) , until he reaches a speed of 3.4 \(\mathrm{m} / \mathrm{s}\) . What is the value of his acceleration (assumed to be constant) during the initial 1.5 -s period?

One afternoon, a couple walks three-fourths of the way around a circular lake, the radius of which is 1.50 \(\mathrm{km} .\) They start at the west side of the lake and head due south to begin with. (a) What is the distance they travel? (b) What are the magnitude and direction (relative to due east) of the couple's displacement?

A car makes a trip due north for three-fourths of the time and due south one- fourth of the time. The average northward velocity has a magnitude of \(27 \mathrm{m} / \mathrm{s},\) and the average southward velocity has a magnitude of 17 \(\mathrm{m} / \mathrm{s}\) . What is the average velocity (magnitude and direction) for the entire trip?

A hot-air balloon is rising upward with a constant speed of 2.50 \(\mathrm{m} / \mathrm{s}\) . When the balloon is 3.00 \(\mathrm{m}\) above the ground, the balloonist accidentally drops a compass over the side of the balloon. How much time elapses before the compass hits the ground?

A sprinter explodes out of the starting block with an acceleration of \(+2.3 \mathrm{m} / \mathrm{s}^{2}\) , which she sustains for 1.2 \(\mathrm{s}\) . Then, her acceleration drops to zero for the rest of the race. What is her velocity \((\mathrm{a})\) at \(t=1.2 \mathrm{s}\) and (b) at the end of the race?
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