A car is parked on a steep incline, making an angle of 37.0° below the horizontal and overlooking the ocean, when its brakes fai
l and it begins to roll. Starting from rest at t = 0, the car rolls down the incline with a constant acceleration of 4.05 m/s2, traveling 46.5 m to the edge of a vertical cliff. The cliff is 30.0 m above the ocean. (a) Find the speed of the car when it reaches the edge of the cliff. m/s (b) Find the time interval elapsed when it arrives there. s (c) Find the velocity of the car when it lands in the ocean. magnitude m/s direction ° below the horizontal (d) Find the total time interval the car is in motion. s (e) Find the position of the car when it lands in the ocean, relative to the base of the cliff. m
1 answer:
Answer:
a) The speed of the car when it reaches the edge of the cliff is 19.4 m/s
b) The time it takes the car to reach the edge is 4.79 s
c) The velocity of the car when it lands in the ocean is 31.0 m/s at 60.2º below the horizontal
d) The total time interval the car is in motion is 6.34 s
e) The car lands 24 m from the base of the cliff.
Explanation:
Please, see the figure for a description of the situation.
a) The equation for the position of an accelerated object moving in a straight line is as follows:
x =x0 + v0 * t + 1/2 a * t²
where:
x = position of the car at time t
x0 = initial position
v0 = initial velocity
t = time
a = acceleration
Since the car starts from rest and the origin of the reference system is located where the car starts moving, v0 and x0 = 0. Then, the position of the car will be:
x = 1/2 a * t²
With the data we have, we can calculate the time it takes the car to reach the edge and with that time we can calculate the velocity at that point.
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
<u>t = 4.79 s </u>
The equation for velocity is as follows:
v = v0 + a* t
Where:
v = velocity
v0 = initial velocity
a = acceleration
t = time
For the car, the velocity will be
v = a * t
at the edge, the velocity will be:
v = 4.05 m/s² * 4.79 s = <u>19.4 m/s</u>
b) The time interval was calculated above, using the equation of the position:
x = 1/2 a * t²
46.5 m = 1/2 * 4.05 m/s² * t²
2* 46.5 m / 4.05 m/s² = t²
t = 4.79 s
c) When the car falls, the position and velocity of the car are given by the following vectors:
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
v =(v0x, v0y + g * t)
Where:
r = position vector
x0 = initial horizontal position
v0x = initial horizontal velocity
t = time
y0 = initial vertical position
v0y = initial vertical velocity
g = acceleration due to gravity
v = velocity vector
First, let´s calculate the initial vertical and horizontal velocities (v0x and v0y). For this part of the problem let´s place the center of the reference system where the car starts falling.
Seeing the figure, notice that the vectors v0x and v0y form a right triangle with the vector v0. Then, using trigonometry, we can calculate the magnitude of each velocity:
cos -37.0º = v0x / v0
(the angle is negative because it was measured clockwise and is below the horizontal)
(Note that now v0 is the velocity the car has when it reaches the edge. it was calculated in a) and is 19,4 m/s)
v0x = v0 * cos -37.0 = 19.4 m/s * cos -37.0º = 15.5 m/s
sin 37.0º = v0y/v0
v0y = v0 * sin -37.0 = 19.4 m/s * sin -37.0 = - 11. 7 m/s
Now that we have v0y, we can calculate the time it takes the car to land in the ocean, using the y-component of the vector "r final" (see figure):
y = y0 + v0y * t + 1/2 * g * t²
Notice in the figure that the y-component of the vector "r final" is -30 m, then:
-30 m = y0 + v0y * t + 1/2 * g * t²
According to our reference system, y0 = 0:
-30 m = v0y * t + 1/2 g * t²
-30 m = -11.7 m/s * t - 1/2 * 9.8 m/s² * t²
0 = 30 m - 11.7 m/s * t - 4.9 m/s² * t²
Solving this quadratic equation:
<u>t = 1.55 s</u> ( the other value was discarded because it was negative).
Now that we have the time, we can calculate the value of the y-component of the velocity vector when the car lands:
vy = v0y + g * t
vy = - 11. 7 m/s - 9.8 m/s² * 1.55s = -26.9 m/s
The x-component of the velocity vector is constant, then, vx = v0x = 15.5 m/s (calculated above).
The velocity vector when the car lands is:
v = (15.5 m/s, -26.9 m/s)
We have to express it in magnitude and direction, so let´s find the magnitude:
To find the direction, let´s use trigonometry again:
sin α = vy / v
sin α = 26.9 m/s / 31.0 m/s
α = 60.2º
(notice that the angle is measured below the horizontal, then it has to be negative).
Then, the vector velocity expressed in terms of its magnitude and direction is:
vy = v * sin -60.2º
vx = v * cos -60.2º
v = (31.0 m/s cos -60.2º, 31.0 m/s sin -60.2º)
<u>The velocity is 31.0 m/s at 60.2º below the horizontal</u>
d) The total time the car is in motion is the sum of the falling and rolling time. This times where calculated above.
total time = falling time + rolling time
total time = 1,55 s + 4.79 s = <u>6.34 s</u>
e) Using the equation for the position vector, we have to find "r final 1" (see figure):
r = (x0 + v0x * t, y0 + v0y * t + 1/2 * g * t²)
Notice that the y-component is 0 ( figure)
we have already calculated the falling time and the v0x. The initial position x0 is 0. Then.
r final 1 = ( v0x * t, 0)
r final 1 = (15.5 m/s * 1.55 s, 0)
r final 1 = (24.0 m, 0)
<u>The car lands 24 m from the base of the cliff.</u>
PHEW!, it was a very complete problem :)
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Answer:
1) f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) the two diameters have the same order of magnitude and are very close to each other
Explanation:
You have some problems in the writing of your exercise, we will try to answer.
1) The equation to be used in geometric optics is the constructor equation
where p and q are the distance to the object and the image, respectively, f is the focal length
* For the normal eye and with presbyopia
the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)
f'₀ = 1.5 cm
this is the focal length for this type of eye
* Eye with myopia
the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm
1 / f = 1/15 + 1 / 1.5
1 / f = 0.733
f = 1.36 cm
this is the focal length for the myopic eye.
In general, the two focal lengths are related
f’₀ / f = 1.5 / 1.36
f’₀ / f = 1.10
The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation
2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit
a sin θ= m λ
the first zero occurs for m = 1, as the angles are very small
tan θ = y / f = sin θ / cos θ
for some very small the cosine is 1
sin θ = y / f
where f is the distance of the lens (eye)
y / f = lam / a
in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor
y / f = 1.22 λ / D
y = 1.22 λ f / D
where D is the diameter of the eye
D = 2R₀
D = 2 0.1
D = 0.2 cm
the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation
* normal eye
the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation
\frac{1}{f} = \frac{1}{p} + \frac{1}{q}
sustitute
= 0.7066
f = 1.415 cm
therefore the diffraction is
y = 1.22 550 10⁻⁹ 1.415 / 0.2
y = 4.75 10⁻⁶ m
this is the radius, the diffraction diameter is
d = 2y
d_normal = 9.49 10⁻⁶ m
* myopic eye
In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm
= 0.733
f = 1.36 cm
diffraction is
y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2
y = 4.56 10-6 m
the diffraction diameter is
d_myope = 2y
d_myope = 9.16 10-6 m
= 9.49 /9.16
\frac{d_{normal}}{d_{myope}} = 1.04
we can see that the two diameters have the same order of magnitude and are very close to each other