Question

An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when a strong wind from the south imparts a constant acceleration of 0.200 m/s^2. If the wind's acceleration lasts for 3.40 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the +x direction and that there are 1609 m in 1 mi. r = 3.25 θ = 20.8° Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a φ = 48.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s^2, find the displacement vector r and the angle of the displacement θ1. Enter the components of the vector, r_x and r_y, and the angle. Assume the time interval is still 3.40 s. r = r_x i + r_y j r_x = ? r_y = ? θ1 = ?

          An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when a strong wind from the south imparts a constant acceleration of 0.200 m/s^2. If the wind's acceleration lasts for 3.40 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the +x direction and that there are 1609 m in 1 mi.
r = 3.25
θ = 20.8°

Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a φ = 48.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s^2, find the displacement vector r and the angle of the displacement θ1. Enter the components of the vector, r_x and r_y, and the angle. Assume the time interval is still 3.40 s.
r = r_x i + r_y j
r_x = ?
r_y = ?
θ1 = ?

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Added by Derek J.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Adi S

09/24/2022

Step 1

- Given: \( 2.00 \, \text{mph} \) - Conversion factor: \( 1 \, \text{mile} = 1609 \, \text{m} \) and \( 1 \, \text{hour} = 3600 \, \text{s} \) \[ 2.00 \, \text{mph} = 2.00 \times \frac{1609 \, \text{m}}{3600 \, \text{s}} \approx 0.894 \, \text{m/s} \] Show more…

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An unsuspecting bird is coasting along in an easterly direction at 2.00 mph when a strong wind from the south imparts a constant acceleration of 0.200 m/s^2. If the wind's acceleration lasts for 3.40 s, find the magnitude r and direction θ (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. Assume that the bird is originally traveling in the +x direction and that there are 1609 m in 1 mi. r = 3.25 θ = 20.8° Now, assume the same bird is moving along again at 2.00 mph in an easterly direction, but this time, the acceleration given by the wind is at a φ = 48.0° angle to the original direction of motion, measured counterclockwise from the easterly direction. If the magnitude of the acceleration is 0.300 m/s^2, find the displacement vector r and the angle of the displacement θ1. Enter the components of the vector, r_x and r_y, and the angle. Assume the time interval is still 3.40 s. r = r_x i + r_y j r_x = ? r_y = ? θ1 = ?

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Transcript

00:02 In this question, it is given that the speed of bird which can be written as u equal to 2 .00 miles per hour, you can be converted to meter per second, so 2 .00 times 1 mile is 1609 meter and 1 hour equal to 3 ,600 second.

00:28 So the speed of the bird can be calculated as 0 .894 meter per second.

00:36 And the acceleration of wind is given as a equal to 0 .200 meter per seconds.

00:47 And the time elapsed, t is given as 3 .40 second.

00:53 So we can find the displacement of bird along the east direction.

00:59 So by using the kinematic equation, x will be equal to ut.

01:05 That can be calculated as by substituting the value 0 .894 meter per second times 3 .40 second can be calculated as 3 .04 meter.

01:20 Similarly, the displacement of bird along the north direction, that is, strong wind imparts from the south.

01:30 So y equal to half a d square.

01:34 We can substitute the value it will become half into acceleration is 0 .200 meter per second square times 3 .40 second whole square can be calculated as 1 .16 meter.

01:50 So the net displacement of the bird can be written as modulus of r equal to x square plus y square we can substitute the value it will become root of 3 .04 meter square plus 1 .16 meter and the displacement can be calculated as 3 .25 meter and we need to find the direction of displacement so direction of displacement 5 equal to tan inverse y by x.

02:32 We can substitute the value so it will become tan inverse 1 .16 meter divided by 3 .04 meter...

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