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A train slows down as it rounds a sharp horizontal turn, slowing from 90.0 km/h to 50.0 km/h in the 15.0 s that it takes to round the bend. The radius of the curve is 150 m. Compute the acceleration at the moment the train speed reaches 50.0 km/h. Assume it continues to slow down at this time at the same rate.

Short Answer

Expert verified
The total magnitude of the acceleration of the train is \(a_{total} = \sqrt{a^2 + a_c^2}\), combining both the tangential and centripetal accelerations.

Step by step solution

01

Convert Speeds to Meters per Second

Given speeds are in kilometers per hour (km/h), so convert them into meters per second (m/s) using the conversion factor \(1\ \text{km/h} = \frac{1}{3.6}\ \text{m/s}\). Thus, initial speed \(u = 90.0\ \text{km/h}\) becomes \(u = \frac{90.0}{3.6}\ \text{m/s}\) and final speed \(v = 50.0\ \text{km/h}\) becomes \(v = \frac{50.0}{3.6}\ \text{m/s}\).
02

Calculate the Train's Change in Velocity

After converting the initial and final velocity to m/s, find the change in velocity \(\Delta v = v - u\).
03

Determine the Average Acceleration

Use the formula for average acceleration \(a = \frac{\Delta v}{\Delta t}\), where \(\Delta v\) is the change in velocity and \(\Delta t = 15.0\ \text{s}\) is the time interval.
04

Convert the Centripetal Acceleration to Vector Form

Knowing that the centripetal acceleration is always directed towards the center of the circular path and given by \(a_c = \frac{v^2}{r}\), where \(v = \frac{50.0}{3.6} \text{m/s}\) at the moment of interest and \(r = 150\ \text{m}\) is the radius of the curve. Convert this into vector form considering the direction towards the center as negative.
05

Combine the Tangential and Centripetal Acceleration

The total acceleration is a combination of tangential and centripetal acceleration. Since the train is slowing down, the tangential acceleration is in the opposite direction of the movement and centripetal acceleration is towards the center, therefore, they are perpendicular to each other. Calculate the magnitude using vector addition: \(a_{total} = \sqrt{a^2 + a_c^2}\).

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Key Concepts

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

Kinematics
Kinematics is a branch of classical mechanics that describes the motion of objects without considering the forces that cause the movement. It involves the analysis of position, velocity, and acceleration in different framings of time. For instance, in physics acceleration problems, kinematics principles help us understand how an object like a train, mentioned in the exercise, changes its speed and direction over time.

To determine the train's acceleration as it rounds a turn, we must analyze its change in velocity within a given time frame, which is a fundamental kinematics concept. This involves calculating both the average linear acceleration, which relates to how fast it slows down on a straight path, and the centripetal acceleration, which is tied to the train's change in direction along a curved path.
Centripetal Acceleration
Centripetal acceleration is the acceleration experienced by an object moving in a circular path, directed towards the center of the circle. This force is responsible for keeping the object in its curved path and is a key topic when discussing physics acceleration problems involving circular motion.

For example, as the train moves around the bend, it experiences centripetal acceleration because its direction is continuously changing, even though its speed at the moment of interest is constant. The formula for centripetal acceleration is \(a_c = \frac{v^2}{r}\), with \(v\) being the velocity of the object and \(r\) being the radius of the circular path. Thus, an increase in speed or a decrease in the radius would result in greater centripetal acceleration.
Velocity Conversion
Velocity conversion is a practical skill necessary for solving kinematics problems, especially when units differ from the standard forms used in equations. In the case of our exercise, the train's speed is initially provided in kilometers per hour (km/h), which is not compatible with the standard unit of meters per second (m/s) required for most kinematic formulas.

To convert from km/h to m/s, you multiply the speed value by \(\frac{1}{3.6}\). This conversion is essential because it ensures that all the values used in subsequent calculations are in consistent units, allowing for correct computation of acceleration. Failure to convert velocity accurately would lead to incorrect results and a misunderstanding of the train's motion.
Average Acceleration
Average acceleration is defined as the change in velocity over the change in time and is expressed mathematically as \(a = \frac{\Delta v}{\Delta t}\). It is a vector quantity, meaning it has both magnitude and direction.

In our problem, calculating the train's average acceleration is crucial to understand how quickly it is slowing down. Because the exercise indicates that the train is decelerating, it's important to recognize that the sign of the acceleration will be opposite to the direction of the train's initial velocity. Remember, a positive or negative sign on acceleration indicates direction, not necessarily an increase or decrease in speed.
Vector Addition
Vector addition is a method used in physics to combine different vectors. Vectors are quantities that have both magnitude and direction, such as velocity and acceleration. To solve complex problems, especially those involving more than one dimension, vector addition is necessary.

In the exercise involving the train, the total acceleration of the train at any given moment is the vector sum of its tangential acceleration (due to its change in speed) and its centripetal acceleration (due to its change in direction). Since these two components of acceleration are perpendicular to each other, we use the Pythagorean theorem to find the resultant acceleration: \(a_{total} = \sqrt{a^2 + a_c^2}\). Understanding vector addition is essential for accurately describing the complete state of motion of the train as it slows down while rounding the bend.

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Most popular questions from this chapter

Your grandfather is copilot of a bomber, flying horizontally over level terrain, with a speed of 275 m/s relative to the ground, at an altitude of 3 000 m. (a) The bombardier releases one bomb. How far will it travel horizontally between its release and its impact on the ground? Neglect the effects of air resistance. (b) Firing from the people on the ground suddenly incapacitates the bombardier before he can call, 鈥淏ombs away!鈥 Consequently, the pilot maintains the plane鈥檚 original course, altitude, and speed through a storm of flak. Where will the plane be when the bomb hits the ground? (c) The plane has a telescopic bomb sight set so that the bomb hits the target seen in the sight at the time of release. At what angle from the vertical was the bomb sight set?

A tire 0.500 m in radius rotates at a constant rate of 200 rev/min. Find the speed and acceleration of a small stone lodged in the tread of the tire (on its outer edge).

The vector position of a particle varies in time according to the expression \(\mathbf{r}=\left(3.00 \hat{\mathbf{i}}-6.00 t^{2} \mathbf{j}\right) \mathrm{m} .\) (a) Find expressions for the velocity and acceleration as functions of time. (b) Determine the particle's position and velocity at \(t=1.00 \mathrm{s}\) .

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 \(t=2.00 \mathrm{~s}\).

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.

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