91Ó°ÊÓ

You are to ride a jet-cycle over a lake, starting from rest at point \(1:\) First, moving at \(30^{\circ}\) north of due east: 1\. Increase your speed at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(6.00 \mathrm{~s}\). 2. With whatever speed you then have, move for \(8.00 \mathrm{~s}\). 3\. Then slow at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(6.00 \mathrm{~s}\). Immediately next, moving due west: 4\. Increase your speed at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) for \(5.00 \mathrm{~s}\). 5\. With whatever speed you then have, move for \(10.0 \mathrm{~s}\). 6\. Then slow at \(0.400 \mathrm{~m} / \mathrm{s}^{2}\) until you stop. In magnitude-angle notation, what then is your average velocity for the trip from point 1 ?

Step by step solution

01

- Calculate Velocity After 6 Seconds

Use the formula for acceleration to find the velocity after increasing speed:\(v_f = v_i + a t\)Initial velocity, \(v_i = 0\), acceleration, \(a = 0.400 \, \text{m/s}^2\), time, \(t = 6.00 \, \text{s}\).Plug in the values:\(v_f = 0 + (0.400 \, \text{m/s}^2)(6.00 \, \text{s}) = 2.40 \, \text{m/s}\).

02

- Calculate Distance Covered in 8 Seconds

Use the formula for distance:\(d = v t\)Velocity, \(v = 2.40 \, \text{m/s}\), time, \(t = 8.00 \, \text{s}\).\(d = (2.40 \, \text{m/s})(8.00 \, \text{s}) = 19.20 \, \text{m}\).

03

- Calculate Final Velocity After Slowing Down

Use the formula for deceleration:\(v_f = v_i - a t\)Initial velocity, \(v_i = 2.40 \, \text{m/s}\), acceleration, \(a = 0.400 \, \text{m/s}^2\), time, \(t = 6.00 \, \text{s}\).Plug in the values:\(v_f = 2.40 \, \text{m/s} - (0.400 \, \text{m/s}^2)(6.00 \, \text{s}) = 0 \, \text{m/s}\).

04

- Calculate Velocity After Moving Due West

Use the acceleration formula:\(v_f = v_i + a t\)Initial velocity, \(v_i = 0\), acceleration, \(a = 0.400 \, \text{m/s}^2\), time, \(t = 5.00 \, \text{s}\).\(v_f = 0 + (0.400 \, \text{m/s}^2)(5.00 \, \text{s}) = 2.00 \, \text{m/s}\).

05

- Calculate Distance Covered in 10 Seconds

Use the formula for distance:\(d = v t\)Velocity, \(v = 2.00 \, \text{m/s}\), time, \(t = 10.0 \, \text{s}\).\(d = (2.00 \, \text{m/s})(10.0 \, \text{s}) = 20.0 \, \text{m}\).

06

- Calculate Distance Covered While Slowing Down

Use the equation for motion under constant acceleration:\(v_f^2 = v_i^2 + 2 a d\)Final velocity, \(v_f = 0\), initial velocity, \(v_i = 2.00 \, \text{m/s}\), acceleration is deceleration here, \(a = -0.400 \, \text{m/s}^2\).Plug in and solve for \(d\):\(0 = (2.00 \, \text{m/s})^2 + 2(-0.400 \, \text{m/s}^2)d\)\(0 = 4.00 \, \text{m}^2/\text{s}^2 - 0.800 \, \text{m/s}^2 \, d\)\(d = 4.00 / 0.800 = 5.00 \, \text{m}\).

07

- Calculate Total Displacement

To find the total displacement, sum the displacements from each part:For each component:- 30° North of East: Use trigonometric components:\(s_x = 2.40 \, \text{m/s} \, \times 6.00 \, \text{s} \, \times \, \cos(30°) + 19.20 \, \text{m} \, \times \, \cos(30°)\)\(s_x = 2.80 \, \text{m} \, + 13.60 \, \text{m}\)\(s_x = 16.40 \, \text{m}\)\(s_y = 2.40 \, \text{m/s} \, \times 6.00 \, \text{s} \, \times \, \sin(30°) + 19.20 \, \text{m} \, \times \, \sin(30°)\)\(s_y = 2.80 \, \text{m} \, \times \, 0.5 + 9.60 \, \text{m}\)\(s_y = 11.20 \, \text{m}\)- Due west: Displacement is negative X direction\(-22.00 \, \text{m}\)\(net_x = 16.40 - 22.00 = -5.60 \, \text{m}\)\(net_y = 11.20 \, \text{m}\)

08

- Calculate Average Velocity

Use the magnitude of the overall displacement and total time to find the average velocity:\[V_{avg} = \frac{D_{total}}{T_{total}}\]\(\sqrt{(-5.60)^2 + (11.20)^2} = 12.52 \text{m}\)Time is 6 + 8 + 6 + 5 + 10 + 5 = 40s.\(V_{avg} = \frac{12.52 \, \text{m}}{40 \, \text{s}} \approx 0.3125 \, \text{m/s}\).To find the angle:\(\theta = \tan^{-1}(\frac{net_y}{net_x}) = \tan^{-1}(\frac{11.2}{5.6}) \approx \tan^{-1}(2) = 63.43^{\circ}\)

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Key Concepts

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

acceleration

In kinematics, acceleration is the rate at which an object's velocity changes over time. It is a vector, meaning it has both magnitude and direction. For instance, in the exercise, the jet-cycle increases its speed at 0.400 m/s² for 6 seconds. When calculating the velocity after this period, you use the formula: \( v_f = v_i + a t \)Here, \(v_i\) is the initial velocity (which is 0 in this case), \(a\) is the acceleration, and \(t\) is the time. Plugging in the values:\( v_f = 0 + (0.400 \, m/s^2)(6.00 \, s) = 2.40 \, m/s \)This result means the jet-cycle reaches a velocity of 2.40 m/s after 6 seconds of acceleration.

velocity

Velocity describes the rate at which an object changes its position. Like acceleration, it is a vector quantity, consisting of both speed and direction. For example, in step 4 of the solution, the jet-cycle moves due west, and its velocity is recalculated after increasing speed for 5 seconds: \( v_f = v_i + a t \)With initial velocity \(v_i = 0\), acceleration \(a = 0.400 \, m/s^2\), and time \(t = 5.00 \, s\), we get: \( v_f = 0 + (0.400 \, m/s^2)(5.00 \, s) = 2.00 \, m/s \).This tells us that, due to a consistent acceleration, the jet-cycle moves west at a final speed of 2.00 m/s.

displacement

Displacement is a vector quantity that reflects the overall change in position of an object. It considers both the distance covered and the direction. In the exercise, the jet-cycle's displacement is calculated in multiple stages. For instance, during the initial part moving 30° north of east, the displacement is broken into components using trigonometry:\( s_x = 2.40 \, m/s \, \times 6.00 \, s \, \times \cos(30°) + 19.20 \, m \, \times \cos(30°) \)and\( s_y = 2.40 \, m/s \, \times 6.00 \, s \, \times \sin(30°) + 19.20 \, m \, \times \sin(30°) \).These calculations show how far the jet-cycle moves in the x- and y-directions, respectively. Summing these components yields the total displacement.

average velocity

Average velocity is the total displacement divided by the total time taken. In the exercise, after computing the net x and y displacements, you find the overall displacement magnitude:\( \sqrt{(-5.60)^2 + (11.20)^2} = 12.52 \, m \).With a total time of 40 seconds, the average velocity is then:\( V_{avg} = \frac{12.52 \, m}{40 \, s} \approx 0.3125 \, m/s \).To find the direction of the average velocity, use the angle calculated via:\( \theta = \tan^{-1}(\frac{net_y}{net_x}) \approx 63.43^{\circ} \).This angle helps to determine the precise direction of travel, making average velocity both a magnitude and a directional vector.

trigonometry

Trigonometry is crucial in solving problems involving angles and distances in kinematics. In the exercise, it's used to decompose movements into their x and y components when traveling at angles. For instance, moving at 30° north of east, we need to find the x and y parts of displacement using cosine and sine functions:\( s_x = 2.40 \, m/s \, \times 6.00 \, s \, \times \cos(30°) \)and\( s_y = 2.40 \, m/s \, \times 6.00 \, s \, \times \sin(30°) \).This allows us to handle the vector quantities separately before summing them to find the total displacement. This method ensures we accurately capture the directionality in complex problems involving multiple direction changes.