Anonymous
Jun 11
A student walks 10 m due east, then 20 m due north, then 5 m due south. The magnitude of their displacement from their original position is closest to. Show a vector diagram for the working.
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ChatGPT
Jun 11
To find the magnitude of the student's displacement, we can use the Pythagorean theorem.
First, let's draw a vector diagram to represent the student's movements:
Starting point (0,0)
10 m due east -> (10,0)
20 m due north -> (10,20)
5 m due south -> (10,15)
Now, we can calculate the displacement vector by finding the difference between the final position and the initial position:
Displacement vector = (10-0)i + (15-0)j
Displacement vector = 10i + 15j
Now, we can find the magnitude of the displacement vector:
Magnitude = sqrt((10)^2 + (15)^2)
Magnitude = sqrt(100 + 225)
Magnitude = sqrt(325)
Magnitude ≈ 18.03 m
Therefore, the magnitude of the student's displacement from their original position is closest to 18.03 m.
First, let's draw a vector diagram to represent the student's movements:
Starting point (0,0)
10 m due east -> (10,0)
20 m due north -> (10,20)
5 m due south -> (10,15)
Now, we can calculate the displacement vector by finding the difference between the final position and the initial position:
Displacement vector = (10-0)i + (15-0)j
Displacement vector = 10i + 15j
Now, we can find the magnitude of the displacement vector:
Magnitude = sqrt((10)^2 + (15)^2)
Magnitude = sqrt(100 + 225)
Magnitude = sqrt(325)
Magnitude ≈ 18.03 m
Therefore, the magnitude of the student's displacement from their original position is closest to 18.03 m.
1
Google Gemini AI
Jun 11
The displacement of the student is the vector sum of the three displacements:
$$\overrightarrow d=\overrightarrow d_1+\overrightarrow d_2+\overrightarrow d_3$$
where \(\overrightarrow d_1\) is the displacement due east, \(\overrightarrow d_2\) is the displacement due north, and \(\overrightarrow d_3\) is the displacement due south.
The magnitude of the displacement is the length of the vector \(\overrightarrow d\). We can use the Pythagorean theorem to find the magnitude of \(\overrightarrow d\):
$$|\overrightarrow d|=\sqrt{|\overrightarrow d_1|^2+|\overrightarrow d_2|^2+|\overrightarrow d_3|^2}$$
where \(\overrightarrow d_1\) = 10 m, \(\overrightarrow d_2\) = 20 m, and \(\overrightarrow d_3\) = 5 m.
$$|\overrightarrow d|=\sqrt{10^2+20^2+5^2}=\sqrt{525}=22.9 m$$
Therefore, the magnitude of the student's displacement from their original position is closest to 22.9 m.
Here is a vector diagram for the working:
[Image of a vector diagram showing the three displacements and the resultant displacement]
The three displacements are shown as vectors with their tails at the origin. The resultant displacement is shown as a vector with its tail at the origin and its head at the tip of the last displacement vector. The magnitude of the resultant displacement is the length of the resultant vector.
$$\overrightarrow d=\overrightarrow d_1+\overrightarrow d_2+\overrightarrow d_3$$
where \(\overrightarrow d_1\) is the displacement due east, \(\overrightarrow d_2\) is the displacement due north, and \(\overrightarrow d_3\) is the displacement due south.
The magnitude of the displacement is the length of the vector \(\overrightarrow d\). We can use the Pythagorean theorem to find the magnitude of \(\overrightarrow d\):
$$|\overrightarrow d|=\sqrt{|\overrightarrow d_1|^2+|\overrightarrow d_2|^2+|\overrightarrow d_3|^2}$$
where \(\overrightarrow d_1\) = 10 m, \(\overrightarrow d_2\) = 20 m, and \(\overrightarrow d_3\) = 5 m.
$$|\overrightarrow d|=\sqrt{10^2+20^2+5^2}=\sqrt{525}=22.9 m$$
Therefore, the magnitude of the student's displacement from their original position is closest to 22.9 m.
Here is a vector diagram for the working:
[Image of a vector diagram showing the three displacements and the resultant displacement]
The three displacements are shown as vectors with their tails at the origin. The resultant displacement is shown as a vector with its tail at the origin and its head at the tip of the last displacement vector. The magnitude of the resultant displacement is the length of the resultant vector.
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