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expeople1 [14]
3 years ago
6

An automobile follows a circular road whose radius is 50 m. Let x and y respectively denote the eastern and northern directions,

with origin at the center of the circle. Suppose the vehicle starts from rest at x = 50 m heading north, and its speed depends on the distance s it travels according to v = 0.5s − 0.0025s 2 , where s is measured in meters and v is in meters per second. It is known that the tires will begin to skid when the total acceleration of the vehicle is 0.6g. Where will the automobile be and how fast will it be going when it begins to skid? Describe the position in terms of the angle of the radial line relative to the x axis.
Physics
1 answer:
IceJOKER [234]3 years ago
7 0

Answer:

The automobile is running at speed of 23.806 meters per second.

Explanation:

From Kinematic we remember that acceleration (a) can be defined by this ordinary differential equation in terms of distance:

a = v\cdot \frac{dv}{ds} (1)

Where:

v - Speed of the automobile, measured in meters per second.

s - Distance travelled by the automobile, measured in meters.

If we know that v = 0.5\cdot s - 0.0025\cdot s^{2}, then the equation of acceleration is:

a = (0.5\cdot s - 0.0025\cdot s^{2})\cdot \left(0.5-0.0050\cdot s\right)

a = s\cdot (0.5-0.0025\cdot s)\cdot (0.5-0.0050\cdot s)

a = s\cdot (0.0025\cdot s - 0.5)\cdot (0.0050\cdot s-0.5)

But distance covered by the vehicle is defined by the following formula:

s = \theta \cdot r (2)

Where:

\theta - Arc angle, measured in radians.

r - Radius, measured in radians.

Then, we expand (1) by means of this result:

a = \theta\cdot r \cdot (0.0025\cdot \theta\cdot r -0.5)\cdot (0.0050\cdot \theta \cdot r-0.5)

a = \theta\cdot r \cdot (1.25\times 10^{-5}\cdot \theta^{2}\cdot r^{2}-3.75\times 10^{-3}\cdot \theta\cdot r +0.25)

a = 1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r

And finally we get the following third order polynomial:

1.25\times 10^{-5}\cdot \theta^{3}\cdot r^{3}-3.75\times 10^{-3}\cdot \theta^{2}\cdot r^{2}+0.25\cdot \theta \cdot r - a = 0 (3)

If we know that r = 50\,m, a = 0.6\cdot g and g = 9.807\,\frac{m}{s^{2}}, then the polynomial becomes into this:

1.5625\cdot \theta^{3}-9.375\cdot \theta^{2} +12.5\cdot \theta - 5.886 = 0 (3b)

This polynomial can be solved analytically by Cardano's Method or by numerical methods. The roots of the polynomial are, respectivelly:

\theta_{1} \approx 4.365\,rad, \theta_{2} \approx 0.818+i\,0.441\,rad, \theta_{3}\approx 0.818 -i\,0.441\,rad, \theta_{4} \approx 1.563\,rad

Both first and fourth roots are physically reasonable solution, but the latter represents the angle where automobile begins to skid <em>first</em>. Then, the automobile begins to skid at an angle of 1.563 radians relative to x axis.

The distance travelled by the automobile is: (r = 50\,m, \theta \approx 1.563\,rad)

s = (1.563\,rad)\cdot (50\,m)

s = 78.15\,m

Lastly, the speed of the automobile at this location is: (s = 78.15\,m)

v = 0.5\cdot s - 0.0025\cdot s^{2} (4)

v = 0.5\cdot (78.15)-0.0025\cdot (78.15)^{2}

v = 23.806\,\frac{m}{s}

The automobile is running at speed of 23.806 meters per second.

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marishachu [46]
  • initial velocity=u=24m/s
  • Acceleration=a=4m/s^2
  • Distance=s=96m
  • Final velocity=v

Using 3rd equation of kinematics

\boxed{\Large{\sf v^2-u^2=2as}}

\\ \Large\sf\longmapsto v^2=u^2+2as

\\ \Large\sf\longmapsto v^2=24^2+2(4)(96)

\\ \Large\sf\longmapsto v^2=576+768

\\ \Large\sf\longmapsto v^2=1344

\\ \Large\sf\longmapsto v=\sqrt{1344}

\\ \Large\sf\longmapsto v=36.6m/s

3 0
3 years ago
21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe
DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, F_f

Since the two forces are in opposite direction, the equation of motion is

F-F_f = ma

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

F - F_f = 0\\F_f = F

which means that the magnitude of the force of friction is also equal to 40 N.

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3 years ago
Two identical loudspeakers are some distance apart. A person stands 5.80 m from one speaker and 3.90 m from the other. What is t
NikAS [45]

Answer:

f = 632 Hz

Explanation:

As we know that for destructive interference the path difference from two loud speakers must be equal to the odd multiple of half of the wavelength

here we know that

\Delta x = (2n + 1)\frac{\lambda}{2}

given that path difference from two loud speakers is given as

\Delta x = 5.80 m - 3.90 m

\Delta x = 1.90 m

now we know that it will have fourth lowest frequency at which destructive interference will occurs

so here we have

\Delta x = 1.90 = \frac{7\lambda}{2}

\lambda = \frac{2 \times 1.90}{7}

\lambda = 0.54 m

now for frequency we know that

f = \frac{v}{\lambda}

f = \frac{343}{0.54} = 632 Hz

7 0
3 years ago
One mole of magnesium (6 × 1023 atoms) has a mass of 24 grams, as shown in the periodic table on the inside front cover of the t
natka813 [3]

This question involves the concepts of density, volume, and mass.

The approximate diameter of a magnesium atom is "3.55 x 10⁻¹⁰ m".

<h3>STEP 1 (FINDING MASS OF INDIVIDUAL ATOM)</h3>

It is given that:

Mass of one mole = 24 grams

Mass of 6 x 10²³ atoms = 24 grams

Mass of 1 atom = \frac{24\ grams}{6\ x\ 10^{23}\ atoms} = 4 x 10⁻²³ grams

<h3>STEP 2 (FINDING VOLUME OF A SINGLE ATOM)</h3>

\rho = \frac{m}{V}\\\\V=\frac{m}{\rho}

where,

  • \rho = density = 1.7 grams/cm³
  • m = mass of single atom = 4 x 10⁻²³ grams
  • V = volume of single atom = ?

Therefore,

V=\frac{4\ x\ 10^{-23}\ grams}{1.7\ grams/cm^3}

V = 2.35 x 10⁻²³ cm³

<h3>STEP 3 (FINDING DIAMETER OF ATOM)</h3>

The atom is in a spherical shape. Hence, its Volume can be given as follows:

V =\frac{\pi d^3}{6}\\\\d=\sqrt[3]{ \frac{6V}{\pi}}\\\\d=\sqrt[3]{ \frac{6(2.35\ x\ 10^{-23}\ cm^3)}{\pi}}

d = 0.355 x 10⁻⁷ cm = 3.55 x 10⁻¹⁰ m

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