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user100 [1]
3 years ago
6

[7 pts] 1. Suppose that a car starts from rest, its engine providing an acceleration of 4 ft/s2 while air resistance provides 0.

1 ft/s2 of deceleration for each ft per s (foot per second) of the car’s velocity. Set up an Initial Value Problem (IVP) for the velocity of the car. Solve the IVP and use the solution to find the car’s limiting velocity. If the car starts at position x(0) = x0 = 0, ft, what position will it be at after 1 2 hour? Be sure to set up your coordinate system and include units in all of your answers.

Physics
1 answer:
kumpel [21]3 years ago
4 0

Answer:

a) Initial Value Problem

dv/dt = 4 - 0.1v

v(0) = 0

b) solution to the IVP

v(t) = 40(1 - e^(-t/10))

c) Limiting velocity

Vo = 40 ft/s

Position of the car after 12 hours

X = 14,390 ft

Explanations:

The complete explanations of each of the sections contained in the question are in the files attached to this solution.

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a car accelerates from rest at 3.6m/s^2. how much time does it need to attain a speed of 7 m/s? awnser in units of s.
Ivahew [28]
Initial velocity: 0
final velocity: 7 m/s
a = 3.6
t = ?
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(7-0)/3.6 = t
t = 1.94 s
6 0
3 years ago
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Describe what it means to view something from a frame of reference.Give an example to illustrate your explanation.
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2 years ago
Three forces act on an object. Two of the forces are at an angle of 100◦to each other and have magnitude 25N and 12N. The third
seraphim [82]

Answer:

F₄ = 29.819 N

Explanation:

Given

F₁ = (- 25*Cos 50° i + 25*Sin 50° j + 0 k) N

F₂ = (12*Cos 50° i + 12*Sin 50° j + 0 k) N

F₃ = (0 i + 0 j + 4 k) N

Then we have

F₁ + F₂ + F₃ + F₄ = 0

⇒   F₄ = - (F₁ + F₂ + F₃)

⇒   F₄ = - ((- 25*Cos 50° i + 25*Sin 50° j) N + (12*Cos 50° i + 12*Sin 50° j) N + (4 k) N) = (13*Cos 50° i - 37*Sin 50° j - 4 k) N

The magnitude of the force will be

F₄ = √((13*Cos 50°)² + (- 37*Sin 50°)² + (- 4)²) N = 29.819 N

6 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
SOMEONE PLEASE HELP!!
Arisa [49]

Hello!

\large\boxed{\text{C. 7,350,000 J}}

Use the equation:

PE = mgh

Where:

m = mass of the object (kg)

g = acceleration due to gravity (≈9.8 m/s)

h = height above ground (m)

Plug the given values into the equation:

PE = 7500 · 9.8 · 100

PE = 7,350,000 Joules.

7 0
3 years ago
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