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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

An air-hockey puck has a model rocket rigidly attached to it. The puck is pushed from one corner along the long side of the \(2.00-\mathrm{m}\) long air- hockey table, with the rocket pointing along the short side of the table, and at the same time the rocket is fired. If the rocket thrust imparts an acceleration of \(2.00 \mathrm{~m} / \mathrm{s}^{2}\) to the puck, and the table is \(1.00 \mathrm{~m}\) wide, with what minimum initial velocity should the puck be pushed to make it to the opposite short side of the table without bouncing off either long side of the table? Draw the trajectory of the puck for three initial velocities: \(vv_{\min } .\) Neglect friction and air resistance.

A circus juggler performs an act with balls that he tosses with his right hand and catches with his left hand. Each ball is launched at an angle of \(75^{\circ}\) and reaches a maximum height of \(90 \mathrm{~cm}\) above the launching height. If it takes the juggler \(0.2 \mathrm{~s}\) to catch a ball with his left hand, pass it to his right hand and toss it back into the air, what is the maximum number of balls he can juggle?

A baseball is thrown with a velocity of \(31.1 \mathrm{~m} / \mathrm{s}\) at an angle of \(\theta=33.4^{\circ}\) above horizontal. What is the horizontal component of the ball's velocity at the highest point of the ball's trajectory?

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A rocket-powered hockey puck is moving on a (frictionless) horizontal air- hockey table. The \(x\) - and \(y\) -components of its velocity as a function of time are presented in the graphs below. Assuming that at \(t=0\) the puck is at \(\left(x_{0}, y_{0}\right)=(1,2)\) draw a detailed graph of the trajectory \(y(x)\).
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