An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the south imparts a constant acceleration of 0.300 m/s2. If the acceleration from the wind lasts for 2.70 s, find the magnitude, r, and direction, θ, of the bird\'s displacement during this time period. (HINT: assume the bird is originally travelling in the x direction and there are 1609 m in 1 mile.)

Well, in this situation, we can use some good old physics to calculate the bird's displacement. So, let's get quacking!

First, we need to convert the bird's speed from mph to m/s. Since there are 1609 m in 1 mile and the bird is going at 4.00 mph, we have:

4.00 mph * 1609 m/1 mile = 6436 m/h

But we need m/s, so let's convert the m/h to m/s:

6436 m/h * 1 h/3600 s = 1.7889 m/s

So, the bird's initial velocity is 1.7889 m/s eastward.

Now, we can calculate the bird's final velocity after the acceleration from the wind. The formula for that is:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the values we know, we have:

v = 1.7889 m/s + (0.300 m/s²)(2.70 s)
v = 1.7889 m/s + 0.810 m/s = 2.5989 m/s

Now, we can calculate the displacement of the bird using the formula:

s = ut + 0.5at²

where s is the displacement. Plugging in the values, we get:

s = (1.7889 m/s)(2.70 s) + 0.5(0.300 m/s²)(2.70 s)²
s = 4.8416 m + 0.3645 m = 5.2061 m

So, the magnitude of the bird's displacement is 5.2061 m. But what about the direction?

To find the direction, we can use trigonometry. We can consider the bird's displacement as a vector, with its horizontal component being the bird's eastward displacement and its vertical component being its northward displacement.

Since the bird is originally moving in the x direction (east), and the wind is acting in the y direction (south), the bird's displacement will have a positive x component (east) and a negative y component (south).

Now, we can use the tangent of the angle to find the direction:

tan(θ) = opposite/adjacent = (southward displacement)/(eastward displacement)

tan(θ) = (-5.2061 m)/0 = undefined

Uh-oh, it seems that the tangent is undefined when the eastward displacement is zero. This means that the bird is blown off course and doesn't have a specific direction during this time period. So, θ is undefined and the bird's displacement is 5.2061 m.

It seems that the bird had a bit of a "fly-tastrophy" with that gusty wind!

To find the magnitude, r, and direction, θ, of the bird's displacement, we can break down the problem into two components: the horizontal (x) component and the vertical (y) component.

First, let's find the horizontal component of the bird's displacement. The bird is initially coasting in an easterly direction at 4.00 mph. We need to convert this speed to m/s:

4.00 mph = 4.00 * (1609 m / 3600 s) ≈ 1.79 m/s

The horizontal component of the bird's displacement can be calculated using the equation:

x = v0*t + (1/2)*a*t^2

where v0 is the initial velocity, a is the acceleration from the wind, and t is the time duration of acceleration.

Plugging in the values:
v0 = 1.79 m/s (the initial velocity of the bird)
a = 0 (since the wind does not affect the bird's horizontal motion)
t = 2.70 s

x = (1.79 m/s * 2.70 s) + (1/2)(0 m/s^2)(2.70 s)^2
= 4.85 m

So, the horizontal component of the bird's displacement is 4.85 m.

Next, let's find the vertical component of the bird's displacement. The constant acceleration from the wind acts in the south direction. Since the bird is initially coasting in the easterly direction, there is no initial vertical component to consider.

The equation for vertical displacement is given by:

y = (1/2)*a*t^2

Plugging in the values:
a = 0.300 m/s^2 (acceleration from the wind)
t = 2.70 s

y = (1/2)(0.300 m/s^2)(2.70 s)^2
= 0.364 m

So, the vertical component of the bird's displacement is 0.364 m.

Now, we can find the magnitude of the bird's displacement using the Pythagorean theorem:

r = sqrt(x^2 + y^2)
= sqrt((4.85 m)^2 + (0.364 m)^2)
≈ 4.86 m

Therefore, the magnitude of the bird's displacement is approximately 4.86 m.

To find the direction, θ, of the bird's displacement, we can use the inverse tangent function:

θ = arctan(y / x)
= arctan(0.364 m / 4.85 m)
= arctan(0.075)
≈ 4.298° (rounded to three decimal places)

So, the direction of the bird's displacement is approximately 4.298°.

To find the magnitude, r, and direction, θ, of the bird's displacement, you need to consider the bird's initial velocity, the acceleration from the wind, and the time for which the acceleration is applied.

First, let's convert the bird's initial velocity from mph to m/s:
Initial velocity = 4.00 mph = (4.00 * 1609 m) / (1 h * 3600 s) ≈ 1.79 m/s

Next, calculate the change in velocity caused by the wind's acceleration:
Change in velocity = acceleration * time = 0.300 m/s^2 * 2.70 s ≈ 0.81 m/s

Now, add the change in velocity to the initial velocity to get the final velocity of the bird:
Final velocity = initial velocity + change in velocity ≈ 1.79 m/s + 0.81 m/s ≈ 2.60 m/s

To find the displacement of the bird, you need to multiply its final velocity by the time:
Displacement = final velocity * time = 2.60 m/s * 2.70 s ≈ 7.02 m

Now, let's find the magnitude, r, of the displacement using the Pythagorean theorem:
r = sqrt(x^2 + y^2)

Here, x represents the displacement in the x-direction, which is the initial velocity * time:
x = initial velocity * time ≈ 1.79 m/s * 2.70 s ≈ 4.85 m

And y represents the displacement in the y-direction caused by the wind's acceleration, which is the change in velocity * time:
y = change in velocity * time ≈ 0.81 m/s * 2.70 s ≈ 2.19 m

Now, substitute the values of x and y into the equation for r:
r = sqrt(4.85^2 + 2.19^2) ≈ sqrt(23.52 + 4.80) ≈ sqrt(28.32) ≈ 5.32 m

So, the magnitude of the bird's displacement is approximately 5.32 m.

To find the direction, θ, of the bird's displacement, you can use the inverse tangent function:
θ = arctan(y / x) = arctan(2.19 m / 4.85 m)

Using a calculator or math software, you can find that arctan(2.19 / 4.85) ≈ 25.25 degrees.

Therefore, the magnitude, r, of the bird's displacement is approximately 5.32 m, and the direction, θ, is approximately 25.25 degrees with respect to the x-direction.

Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79

m/s

To solve the problem, you need to create a formula with a position as a function of time for the bird(x) and the wind(y), and then use R = sqrt(Rx^2 + Ry^2) to find the resultant magnitude and the inverse tangent of y/x

R = Ri(initial) + Vi*t + 1/2(Ri)t^2

Rx = 0 + 1.79(2.7) + 1/2(0)(2.7)^2 = 4.833

Vi for the wind is 0 because it is with respect to the bird for the y coordinate
Ry = 0 + 0(3.7) + 1/2(0.3)(3.7)^2 = 2.054

R= sqrt(4.833^2 + 2.054^2) = 5.251 meters

A = arctan(2.054/4.833) = 23.025 degrees

Vb = 4mi/h * 1609m/mi * 1h/3600s = 1.79

m/s.

Vw = a*t = 0.30 * 2.7 = 0.81 m/s.

Vr. = sqrt(Vb^2+Vw^2) = 1.96 m/s. = Resultant velocity.
D = Vr*t = 1.96*2.7 = 5.29 m.

Tan A = Y/X = Vw/Vb = 0.81/1.79=0.45251
A = 24.3o = Direction.