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Chapter 3: Problem 5

A jet plane is flying at a constant altitude. At time \(t_{1}=0,\) it has components of velocity \(v_{x}=90 \mathrm{~m} / \mathrm{s}, v_{y}=110 \mathrm{~m} / \mathrm{s} .\) At time \(t_{2}=30.0 \mathrm{~s}\) the components are \(v_{x}=-170 \mathrm{~m} / \mathrm{s}, v_{y}=40 \mathrm{~m} / \mathrm{s}\). (a) Sketch the velocity vectors at \(t_{1}\) and \(t_{2} .\) How do these two vectors differ? For this time interval calculate (b) the components of the average acceleration, and (c) the magnitude and direction of the average acceleration.

Short Answer

Expert verified
The velocity vectors at \(t_{1}\) and \(t_{2}\) differ both in direction and magnitude. The components of the average acceleration are -8.67 m/s虏 in the x-direction and -2.33 m/s虏 in the y-direction. The magnitude of the average acceleration is 9.07 m/s虏 and its direction is 195.1掳 from the positive x-axis.

Step by step solution

01

Sketch Velocity vectors

To sketch the velocity vectors at \(t_{1}\) and \(t_{2}\) begin by drawing a coordinate system. Mark the vectors from the origin \(v_{1}\) and \(v_{2}\) respectively where \( v_{1}\) and \(v_{2}\) correspond to the velocity at \(t_{1} = 0\) and \(t_{2} = 30.0 s\) respectively. Remember, the vector's direction is represented by its angle with the x-axis and the magnitude is depicted by the length of the arrow. Your \(v_{1}\) will point to the first quadrant with coordinates (90,110), while \(v_{2}\) will point towards the second quadrant with coordinates (-170, 40). The difference between these two vectors is given by the vector sum \(v_{2} - v_{1}\).
02

Calculate the components of average acceleration

The average acceleration \(\vec{a}\) can be found using the formula \(\vec{a} = \frac{\Delta \vec{v}}{\Delta t}\) where \(\Delta \vec{v} = \vec{v_{2}} - \vec{v_{1}}\). In the x-direction, this gives \(a_{x} = \frac{v_{2x} - v_{1x}}{t_{2} - t_{1}} = \frac{-170 m/s - 90 m/s}{30.0 s - 0} = -8.67 m/s^2\)In the y-direction, this results in \(a_{y} = \frac{v_{2y} - v_{1y}}{t_{2} - t_{1}} = \frac{40 m/s - 110 m/s}{30.0 s - 0} = -2.33 m/s^2\)
03

Calculate the magnitude of the average acceleration

The magnitude of the average acceleration \(|\vec{a}|\) can be obtained by the Pythagorean theorem: \(|\vec{a}| = \sqrt{{a_x}^2 + {a_y}^2} = \sqrt{(-8.67 m/s^2)^2 + (-2.33 m/s^2)^2} = 9.07 m/s^2\)
04

Calculate the direction of the average acceleration

The direction of the average acceleration can be calculated using trigonometry. The angle \(胃\) with the x-axis is given by \(tan 胃 = \frac{a_y}{a_x}\), so \(胃 = atan(\frac{-2.33}{-8.67}) = 15.1掳\). Adjust this angle according to the appropriate quadrant of \(a\), so the direction of \(\vec{a}\) is 195.1掳 from the positive x-axis (360掳 - 165掳 = 195.1掳).

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Key Concepts

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

Velocity Vectors
Velocity vectors represent the speed and direction of an object in motion. In our exercise, a jet plane's velocity vectors at two different timestamps are analyzed. A velocity vector consists of two components: the x-component (horizontal) and the y-component (vertical). At time \(t_1 = 0\), the velocity vector \(v_1\) has components \((90 \text{ m/s}, 110 \text{ m/s})\), indicating motion towards the first quadrant. At \(t_2 = 30.0 \text{ s}\), the velocity vector \(v_2\) changes to \((-170 \text{ m/s}, 40 \text{ m/s})\), pointing towards the second quadrant.
The direction of a vector is illustrated by the angle it makes with the positive x-axis, and its magnitude, or size, corresponds to the length of the vector arrow. To understand how these vectors change over time, consider both the change in magnitude and direction from \(v_1\) to \(v_2\).
Understanding velocity vectors helps predict the path and speed of moving objects, which is crucial in fields like aviation and navigation.
Components of Velocity
The components of velocity break down a vector into its horizontal and vertical parts. These components are vital for calculating changes in motion over time. For the jet, at \(t_1\), the components are \(v_x = 90 \text{ m/s}\) and \(v_y = 110 \text{ m/s}\). At \(t_2\), these shift to \(v_x = -170 \text{ m/s}\) and \(v_y = 40 \text{ m/s}\).
Splitting a vector into components allows us to analyze each dimension separately, making calculations more manageable. This technique helps in applying mathematical operations like vector addition or subtraction by dealing with each component individually. This way, one can derive the total effect of multiple forces acting on an object in two-dimensional space.
Analyzing the components can thus help us determine the rate of change along each axis independently, which is particularly important in physics problems involving motion.
Magnitude and Direction of Acceleration
Acceleration measures how quickly velocity changes over time and involves both magnitude and direction. To compute the magnitude, use the Pythagorean theorem: \[|\vec{a}| = \sqrt{a_x^2 + a_y^2}\]For our jet plane, the x-component of average acceleration is computed as \(a_x = \frac{-170 \text{ m/s} - 90 \text{ m/s}}{30.0 \text{ s}} = -8.67 \text{ m/s}^2\).Similarly, for the y-component: \(a_y = \frac{40 \text{ m/s} - 110 \text{ m/s}}{30.0 \text{ s}} = -2.33 \text{ m/s}^2\).
Using these, the magnitude of the average acceleration is \(9.07 \text{ m/s}^2\).
The direction is found using the tangent function: \( an \theta = \frac{a_y}{a_x}\).Hence, \(\theta\) is calculated as \(15.1^\circ\). Because the acceleration lies in the third quadrant, adjust by adding 180掳 to find it is \(195.1^\circ\) from the positive x-axis. This indicates the overall direction in which the velocity vector is changing.
Understanding both the magnitude and direction of acceleration is essential for predicting future motion paths for objects, aiding better navigation and control.

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Most popular questions from this chapter

In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path \(90.0 \mathrm{~m}\) above the ground and has a speed of \(64.0 \mathrm{~m} / \mathrm{s}(143 \mathrm{mi} / \mathrm{h}),\) at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.

A dog running in an open field has components of velocity \(v_{x}=2.6 \mathrm{~m} / \mathrm{s}\) and \(v_{y}=-1.8 \mathrm{~m} / \mathrm{s}\) at \(t_{1}=10.0 \mathrm{~s} .\) For the time interval from \(t_{1}=10.0 \mathrm{~s}\) to \(t_{2}=20.0 \mathrm{~s},\) the average acceleration of the dog has magnitude \(0.45 \mathrm{~m} / \mathrm{s}^{2}\) and direction \(31.0^{\circ}\) measured from the \(+x\) -axis toward the \(+y\) -axis. At \(t_{2}=20.0 \mathrm{~s},\) (a) what are the \(x\) - and \(y\) -components of the dog's velocity? (b) What are the magnitude and direction of the dog's velocity? (c) Sketch the velocity vectors at \(t_{1}\) and \(t_{2}\). How do these two vectors differ?

A toy rocket is launched with an initial velocity of \(12.0 \mathrm{~m} / \mathrm{s}\) in the horizontal direction from the roof of a \(30.0-\mathrm{m}\) -tall building. The rocket's engine produces a horizontal acceleration of \(\left(1.60 \mathrm{~m} / \mathrm{s}^{3}\right) t,\) in the same direction as the initial velocity, but in the vertical direction the acceleration is \(g\), downward. Ignore air resistance. What horizontal distance does the rocket travel before reaching the ground?

A \(76.0 \mathrm{~kg}\) rock is rolling horizontally at the top of a vertical cliff that is \(20 \mathrm{~m}\) above the surface of a lake (Fig. \(\mathrm{P} 3.65\) ). The top of the vertical face of a dam is located \(100 \mathrm{~m}\) from the foot of the cliff, with the top of the dam level with the surface of the water in the lake. A level plain is \(25 \mathrm{~m}\) below the top of the dam. (a) What must be the minimum speed of the rock just as it leaves the cliff so that it will reach the plain without striking the dam? (b) How far from the foot of the dam does the rock hit the plain?

An object is projected with initial speed \(v_{0}\) from the edge of the roof of a building that has height \(H .\) The initial velocity of the object makes an angle \(\alpha_{0}\) with the horizontal. Neglect air resistance. (a) If \(\alpha_{0}\) is \(90^{\circ}\), so that the object is thrown straight up (but misses the roof on the way down), what is the speed \(v\) of the object just before it strikes the ground? (b) If \(\alpha_{0}=-90^{\circ}\), so that the object is thrown straight down, what is its speed just before it strikes the ground? (c) Derive an expression for the speed \(v\) of the object just before it strikes the ground for general \(\alpha_{0}\). (d) The final speed \(v\) equals \(v_{1}\) when \(\alpha_{0}\) equals \(\alpha_{1}\). If \(\alpha_{0}\) is increased, does \(v\) increase, decrease, or stay the same?
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