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Chapter 3: Problem 33

What is the magnitude of an object's average velocity if an object moves from a point with coordinates \(x=2.0 \mathrm{~m}\) \(y=-3.0 \mathrm{~m}\) to a point with coordinates \(x=5.0 \mathrm{~m}, y=-9.0 \mathrm{~m}\) in a time interval of \(2.4 \mathrm{~s} ?\)

Short Answer

Expert verified
Answer: The magnitude of the object's average velocity is approximately 1.85 m/s.

Step by step solution

01

Calculate the displacement vector between the two points

The displacement vector is the difference between the final and initial positions. Given the initial position coordinates (\(x_1 = 2.0 \mathrm{~m}, y_1 = -3.0 \mathrm{~m}\)) and final position coordinates (\(x_2 = 5.0 \mathrm{~m}, y_2 = -9.0 \mathrm{~m}\)), the displacement vector \(\vec{d}\) can be calculated as follows: \(\vec{d} = \begin{pmatrix} x_2 - x_1 \\ y_2 - y_1 \end{pmatrix} = \begin{pmatrix} 5.0 - 2.0 \\ -9.0 - (-3.0) \end{pmatrix} = \begin{pmatrix} 3.0 \\ -6.0 \end{pmatrix} \mathrm{~m}\)
02

Calculate the magnitude of the displacement vector

Now we need to find the magnitude of the displacement vector. For a 2D vector \(\vec{d} = \begin{pmatrix} x \\ y \end{pmatrix}\), the magnitude \(|\vec{d}|\) is given by the formula: \(|\vec{d}| = \sqrt{x^2 + y^2}\) Applying the formula to our displacement vector, we get: \(|\vec{d}| = \sqrt{(3.0)^2 + (-6.0)^2} = \sqrt{9 + 36} = \sqrt{45} = 3\sqrt{5} \mathrm{~m}\)
03

Calculate the magnitude of the average velocity

To find the magnitude of the average velocity \(v_{avg}\), we will divide the magnitude of the displacement by the time interval \(\Delta t = 2.4 \mathrm{~s}\). Therefore, the magnitude of the average velocity is: \(v_{avg} = \frac{|\vec{d}|}{\Delta t} = \frac{3\sqrt{5} \mathrm{~m}}{2.4 \mathrm{~s}} \approx 1.85 \mathrm{~m/s}\) So, the magnitude of the object's average velocity is approximately \(1.85 \mathrm{~m/s}\).

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Most popular questions from this chapter

In a proof-of-concept experiment for an antiballistic missile defense system, a missile is fired from the ground of a shooting range toward a stationary target on the ground. The system detects the missile by radar, analyzes in real time its parabolic motion, and determines that it was fired from a distance \(x_{0}=5000 \mathrm{~m}\), with an initial speed of \(600 \mathrm{~m} / \mathrm{s}\) at a launch angle \(\theta_{0}=20^{\circ} .\) The defense system then calculates the required time delay measured from the launch of the missile and fires a small rocket situated at \(y_{0}=500 \mathrm{~m}\) with an initial velocity of \(v_{0} \mathrm{~m} / \mathrm{s}\) at a launch angle \(\alpha_{0}=60^{\circ}\) in the \(y z\) -plane, to intercept the missile. Determine the initial speed \(v_{0}\) of the intercept rocket and the required time delay.

You are walking on a moving walkway in an airport. The length of the walkway is \(59.1 \mathrm{~m}\). If your velocity relative to the walkway is \(2.35 \mathrm{~m} / \mathrm{s}\) and the walkway moves with a velocity of \(1.77 \mathrm{~m} / \mathrm{s}\), how long will it take you to reach the other end of the walkway?

Your friend's car is parked on a cliff overlooking the ocean on an incline that makes an angle of \(17.0^{\circ}\) below the horizontal. The brakes fail, and the car rolls from rest down the incline for a distance of \(29.0 \mathrm{~m}\) to the edge of the cliff, which is \(55.0 \mathrm{~m}\) above the ocean, and, unfortunately, continues over the edge and lands in the ocean. a) Find the car's position relative to the base of the cliff when the car lands in the ocean. b) Find the length of time the car is in the air.

You want to cross a straight section of a river that has a uniform current of \(5.33 \mathrm{~m} / \mathrm{s}\) and is \(127, \mathrm{~m}\) wide. Your motorboat has an engine that can generate a speed of \(17.5 \mathrm{~m} / \mathrm{s}\) for your boat. Assume that you reach top speed right away (that is, neglect the time it takes to accelerate the boat to top speed). a) If you want to go directly across the river with a \(90^{\circ}\) angle relative to the riverbank, at what angle relative to the riverbank should you point your boat? b) How long will it take to cross the river in this way? c) In which direction should you aim your boat to achieve minimum crossing time? d) What is the minimum time to cross the river? e) What is the minimum speed of your boat that will still enable you to cross the river with a \(90^{\circ}\) angle relative to the riverbank?

A cannon is fired from a hill \(116.7 \mathrm{~m}\) high at an angle of \(22.7^{\circ}\) with respect to the horizontal. If the muzzle velocity is \(36.1 \mathrm{~m} / \mathrm{s}\), what is the speed of a 4.35 -kg cannonball when it hits the ground \(116.7 \mathrm{~m}\) below?
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