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Chapter 4: Problem 44

A bolt drops from the ceiling of a train car that is accelerating northward at a rate of \(2.50 \mathrm{m} / \mathrm{s}^{2} .\) What is the acceleration of the bolt relative to \((\mathrm{a})\) the train car? (b) the Earth?

Short Answer

Expert verified
The acceleration of the bolt relative to the train car is 9.8 m/s^2 downward. The total acceleration of the bolt relative to the Earth is approximately 10.11 m/s^2 at an angle pointing northward and downward.

Step by step solution

01

Identify the acceleration of the bolt relative to the train car

Since the bolt is falling inside the train car and is initially at rest relative to it, the acceleration of the bolt relative to the train car is only due to the force of gravity. Therefore, the acceleration of the bolt relative to the train car is the acceleration due to gravity, which is approximately 9.8 m/s^2 downward.
02

Determine the total acceleration relative to the Earth

To find the total acceleration of the bolt relative to the Earth, we need to consider both the acceleration of the train and the acceleration due to gravity. The bolt experiences the same northward acceleration as the train car, 2.50 m/s^2, plus the downward acceleration due to gravity. These two accelerations are perpendicular to each other and combine vectorially to give the total acceleration relative to the Earth.
03

Calculate the magnitude of the total acceleration vector

The total acceleration is a vector with components of 2.50 m/s^2 northward and 9.8 m/s^2 downward. These two components can be combined using the Pythagorean theorem to find the magnitude of the total acceleration: \(a = \sqrt{a_{\text{north}}^2 + a_{\text{gravity}}^2}\). By substituting the known values, we get \(a = \sqrt{(2.50)^2 + (9.8)^2} = \sqrt{6.25 + 96.04} = \sqrt{102.29} \approx 10.11 \mathrm{m/s^2}\).

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Key Concepts

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

Relative Acceleration
Understanding the concept of relative acceleration is crucial in physics as it helps us analyze movement from different reference frames. In our exercise, the bolt's motion is perceived differently when viewed from the train car versus from the Earth.

Relative to the train car, the acceleration of the bolt is only due to the gravitational pull towards the Earth. Despite the train's own acceleration, inside the train, objects appear to fall straight down because it's the only acceleration acting on them in that frame. However, an observer on the ground would see the bolt following a parabolic path, as it is influenced by both the train's acceleration and gravity simultaneously. This is why calculating relative acceleration is a two-step process, considering each reference frame individually.
Acceleration due to Gravity
The acceleration due to gravity, approximately 9.8 m/s虏 on Earth's surface, is a constant vector pointing downwards towards the center of the Earth. In our problem, when the bolt falls from the ceiling of the accelerating train car, it does so because of gravity's pull.

In all situations, whether an object is in free fall or inside a moving vehicle, gravity's acceleration remains consistent and unchanged. It's an essential factor in the calculations for the bolt's overall acceleration from Earth's perspective. Gravity is also a central topic when discussing weight, orbits, and the behavior of projectiles, making it a foundational concept in physics.
Vector Addition in Physics
Vector addition is a method used when dealing with quantities that have both magnitude and direction. In our scenario, the acceleration of the train and the acceleration due to gravity are represented as vectors and must be combined to find the total acceleration relative to the Earth.

Since these two vectors are perpendicular (northward and downward), we can use the Pythagorean theorem to find the resultant vector's magnitude. This vector addition helps us visualize complex motion as a combination of simpler, one-dimensional movements, which is a powerful tool in solving physics problems.
Pythagorean Theorem in Physics
The Pythagorean theorem plays a significant role in physics, particularly when calculating the resultant of perpendicular vectors. This theorem states that in a right-angled triangle, the square of the hypotenuse's length (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides.

In our exercise, once we've determined that the acceleration vectors of the train and gravity are perpendicular, we can treat them as sides of a right triangle with the hypotenuse being the total acceleration. By applying the Pythagorean theorem, we find the magnitude of the bolt's overall acceleration relative to the Earth, which is crucial in understanding the complete motion of the bolt.

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Most popular questions from this chapter

The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a package directly across the river, but you can swim only at 1.50 m/s. (a) If you choose to minimize the time you spend in the water, in what direction should you head? (b) How far downstream will you be carried? (c) What If? If you choose to minimize the distance downstream that the river carries you, in what direction should you head? (d) How far downstream will you be carried?

A fish swimming in a horizontal plane has velocity \(\mathbf{v}_{i}=(4.00 \hat{\mathbf{i}}+1.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) at a point in the ocean where the position relative to a certain rock is \(\mathbf{r}_{i}=(10.0 \hat{\mathbf{i}}-4.00 \hat{\mathbf{j}}) \mathrm{m} .\) After the fish swims with constant acceleration for 20.0 \(\mathrm{s}\) its velocity is \(\mathbf{v}=(20.0 \hat{\mathbf{i}}-5.00 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s}\) . (a) What are the components of the acceleration? (b) What is the direction of the acceleration with respect to unit vector \(\hat{\mathbf{i}}\) ? (c) If the fish maintains constant acceleration, where is it at \(t=25.0 \mathrm{s}\) , and in what direction is it moving?

A golf ball is hit off a tee at the edge of a cliff. Its x and y coordinates as functions of time are given by the following expressions: $$\begin{array}{c}{x=(18.0 \mathrm{m} / \mathrm{s}) t} \\ {\text { and } \quad y=(4.00 \mathrm{m} / \mathrm{s}) t-\left(4.90 \mathrm{m} / \mathrm{s}^{2}\right) t^{2}}\end{array}$$ (a) Write a vector expression for the ball's position as a function of time, using the unit vectors \(\hat{\mathbf{i}}\) and \(\hat{\mathbf{j}}\) . By taking derivatives, obtain expressions for (b) the velocity vector v as a function of time and (c) the acceleration vector a as a function of time. Next use unit- vector notation to write expressions for (d) the position, (e) the velocity, and (f) the acceleration of the golf ball, all at t " 3.00 s.

The determined coyote is out once more in pursuit of the elusive roadrunner. The coyote wears a pair of Acme jet-powered roller skates, which provide a constant horizontal acceleration of 15.0 \(\mathrm{m} / \mathrm{s}^{2}\) (Fig. P4.65). The coyote starts at rest 70.0 \(\mathrm{m}\) from the brink of a cliff at the instant the roadrunner zips past him in the direction of the cliff. (a) If the roadrunner moves with constant speed, determine the minimum speed he must have in order to reach the cliff before the coyote. At the edge of the cliff, the roadrunner escapes by making a sudden turn, while the coyote continues straight ahead. His skates remain horizontal and continue to operate while he is in flight, so that the coyote鈥檚 acceleration while in the air is \((15.0 \hat{\mathrm{i}}-9.80 \hat{\mathrm{j}}) \mathrm{m} / \mathrm{s}^{2} .\) (b) If the cliff is 100 \(\mathrm{m}\) above the flat floor of a canyon, determine where the coyote lands in the canyon. (c) Determine the components of the coyote's impact velocity.

A particle initially located at the origin has an acceleration of \(\mathbf{a}=3.00 \hat{\mathbf{j}} \mathrm{m} / \mathrm{s}^{2}\) and an initial velocity of \(\mathbf{v}_{i}=500 \hat{\mathbf{i}} \mathrm{m} / \mathrm{s}\). Find (a) the vector position and velocity at any time \(t\) and (b) the coordinates and speed of the particle at \(t=2.00 \mathrm{~s}\).
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