Answer: 4100 Mpc
Explanation:
Since H o = 70 km/s/Mpc
Redshift z = 5.82
Recessional velocity vr = 287,000 km/s
Then, the distance to the galaxy in light years will be:
= Recessional velocity / H o
= 287000 / 70
= 4100 Mpc
A car moves forward up a hill at 12 m/s with a uniform backward acceleration of 1.6 m/s2. What is its displacement after 6 s?
Answer:
The displacement of the car after 6s is 43.2 m
Explanation:
Given;
velocity of the car, v = 12 m/s
acceleration of the car, a = -1.6 m/s² (backward acceleration)
time of motion, t = 6 s
The displacement of the car after 6s is given by the following kinematic equation;
d = ut + ¹/₂at²
d = (12 x 6) + ¹/₂(-1.6)(6)²
d = 72 - 28.8
d = 43.2 m
Therefore, the displacement of the car after 6s is 43.2 m
When four people with a combined mass of 310 kg sit down in a 2000-kg car, they find that their weight compresses the springs an additional 0.90 cm. (a) what is the effective force constant of the springs? in N/m (b) The four people get out of the car and bounce it up and down. What is the frequency of the car's vibration?
Answer:
Explanation:
F=kx
x=F/k
F=2000 kg
x=100 cm=9*10^-3
effective spring constant=k=F/x
k=2000/9*10^-3=2.2*10^-5
now frequency
f=1/2π√k/m
f=1/2*3.14√2.2*10^-5/310
f=1/6.28√7.097*10^-8
f=1/6.28*2.7*10^-4
f=0.16*2.7*10^-4
f=4.32*10^-5
The effective spring constant of the springs is 33755.55 N/m.
The frequency of the car's vibration is 2.07 Hz.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
Weight of the four people: F = 310 × 9.80 N = 3038 Newton.
The additional compression of the spring: x = 0.90 cm = 0.90 × 10⁻² m.
Hence, the effective spring constant of the springs: k= force/compression
= 3038 N/0.90 × 10⁻² m
= 33755.55 N/m.
The frequency of the car's vibration is: f = 1/2π√(k/m)
=1/2π√(33755.55/2000)
= 2.07 Hz.
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An object with a mass of 3.0 kg has a
force of 9.0 newtons applied to it. What
is the resulting acceleration of the
object?
[tex] \LARGE{ \underline{ \tt{Required \: answer:}}}[/tex]
We have:
Mass of the object = 3 kgForce on the object = 9 NWe need to find:
Resulting accleration of the object?Solution:
According to Newton's 2nd law of motion, or quantitative measure of Force:
Force = Mass × AcclerationUsing this,
➝ F = ma
➝ 9N = 3 kg × a
➝ a = 9/3 m/s²
➝ a = 3 m/s²
Hence,
The resulting accleration of the object is 3 m/s². And we are done! :D⛱️ [tex] \large{ \blue{ \bf{FadedElla}}}[/tex]
Consider a block sliding down a ramp whose motion is opposed by frictional forces. The total energy of this system is modeled by the equation:
Etotal = 1/2mv^2 + mgh + Ff(f is underscore)d
Which part of the equation represents the amount of energy converted to thermal energy?
A. mg
B. Ffd
C. mgh
D. 1/2 mv^2
Answer:
Energy Flows Quick check answers:
1. Ffd.
2. The kinetic energy decreases, and gravitational potential energy increases.
3. The internal energy of the system increases.
4. KEbox= Etotal-mgh
5. Etotal = 1/2m1(v1)^2+1/2m^2(v2)^2+U
The part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
The given equation for the total energy of a system;
[tex]E_{total} = \frac{1}{2} mv^2 \ +\ mgh\ + \ \ F_fd[/tex]
The definition of the various terms in the energy equation is given as;
[tex]E_{total}[/tex]: this is the total mechanical energy of the system[tex]\frac{1}{2} mv^2[/tex]: this is the kinetic energy of the system[tex]mgh[/tex]: this is the potential energy of the system[tex]F_f d[/tex]: this is the energy lost due to friction.The energy lost due to friction is equal to the energy converted to thermal energy.
Thus, the part of the equation that represents the amount of energy converted to thermal energy is [tex]F_f d[/tex].
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Write Radar gun summary
Answer:
A radar gun is a device for measuring the speed of moving objects. ... The radar gun is a Doppler radar unit that can be static, vehicle-mounted or hand-held. It measures the
Explanation:
someone help please
waves disturb ____, but do not transmit it.
a. energy
b. matter
c. sound
d. none of the above
Answer:
b. matter
Explanation:
Waves disturb matter but do not transmit it.
Waves are disturbances that transmit energy from one point to another. Although they cause disturbances, they do not transfer the matters in the medium.
Energy is propagated by a wave. When for example, sound waves are produced, the disturbance is propagated via particle - particle interaction But after the wave train moves, the particles remain.A household refrigerator consumes electrical energy at the rate of 200 W. lf electricity costs 5 k per kWh, calculate the cost of operating the appliance for 30 days
Answer:
= 720000 [k]
Explanation:
The cost is equal to 5 [$/kW-h], kilowatt per hour, this value should be multiplied by the power, and then by the time.
[tex]5[\frac{k}{kw*h}]*200[w]*30[day]*24[\frac{h}{day} ][/tex]
= 720000 [k]
What is magnet made of
Answer:
metals like iron or nickel
Explanation:
A plane is flying due west at 34 m/s. It encounters a wind blowing at 19 m/s south. Find the resultant veloci
Answer:
The resultant velocity has a magnitude of 38.95 m/s
Explanation:
Vector Addition
Given two vectors defined as:
[tex]\vec v_1=(x_1,y_1)[/tex]
[tex]\vec v_2=(x_2,y_2)[/tex]
The sum of the vectors is:
[tex]\vec v=(x_1+x_2,y_1+y_2)[/tex]
The magnitude of a vector can be calculated by
[tex]d=\sqrt{x^2+y^2}[/tex]
Where x and y are the rectangular components of the vector.
We have a plane flying due west at 34 m/s. Its velocity vector is:
[tex]\vec v_1=(-34,0)[/tex]
The wind blows at 19 m/s south, thus:
[tex]\vec v_2=(0,-19)[/tex]
The sum of both velocities gives the resultant velocity:
[tex]\vec v =(-34,-19)[/tex]
The magnitude of this velocity is:
[tex]d=\sqrt{(-34)^2+(-19)^2}[/tex]
[tex]d=\sqrt{1156+361}=\sqrt{1517}[/tex]
d = 38.95 m/s
The resultant velocity has a magnitude of 38.95 m/s
List at least 4 aspects to evaluate the quality of an internet site
Answer:
authority, accuracy, objectivity, currency, coverage, and appearance.
Explanation:
There are six (6) criteria that should be applied when evaluating any Web site. or each criterion, there are several questions to be asked. The more questions you can answer "yes", the more likely the Web site is one of quality.
it is the question 12 part okay
Answer:
Yeah it's ok I think. Also, I can't see the answer you gave so maybe updating the question would be nice.
How far out from the sun is the rock line?
Answer:
5 million miles
Explanation:
A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?
Answer:
599.245km/hr
Explanation:
A plane is heading due west and climbing at the rate of 80 km/hr. If its airspeed is 540 km/hr and there is a wind blowing 80 km/hr to the northwest, what is the ground speed of the plane?
We solve the above question using vectors
In vector form Air speed is -540i + 0j Wind speed is (-80/√2)i + (80/√2)j
Vector notation wind speed is given as: -56.5685 i + 56.5685j
The vector for the ground speed of the plane =
-540i + 0j -56.5685i + 56.5685j
= -596.56854249i + 56.5685j
The the ground speed of the plane √[(596.56854249)² + (56.5685)²]
= √359094.021081 km/hr
= 599.24454197 km/hr
Approximately
= 599.245km/hr
David Wetterman drops a 5 kg watermelon from the top of a 30 m building. What is the velocity of the watermelon as it smashes
into the ground (neglecting air resistance)?
-(1)
A)
24.25 m/s
B)
32.45 m/s
C)
60 m/s
D)
588 m/s
Answer:
A. 24.25 m/s
Explanation:
velocity = [tex]\sqrt{2 * g * d}[/tex]
velocity = sqr 2 * 9.8 * 30 = sqr 588 = 24.25 m/s
The velocity of the watermelon as it smashes into the ground will be 24.2 m/s
State the third equation of motion?
The third equation of motion is -
v² - u² = 2aS
Given David Wetterman drops a 5 kg watermelon from the top of a 30 m building.
Height of building [S] = 30 m
Mass of watermelon [M] = 5 Kg
Initial velocity [v] = 0 m/s
acceleration [g] = 9.8 m/s²
Using the third equation of motion -
v² - u² = 2aS
v² = 2aS
v² = 2 x 9.8 x 30
v² = 588
v = 24.2 m/s
Therefore, the velocity of the watermelon as it smashes into the ground will be 24.2 m/s.
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I want to know about the inventions caused due to rain. Like the Benjamin Franklin's Lightning Conductor. I have to make a chart.
Acceleration is sometimes expressed in multiples of g, where g = 9.8 m/s^2 is the magnitude of the acceleration due to the earth's gravity. In a test crash, a car's velocity goes from 26 m/s to 0 m/s in 0.15 s. How many g's would be experienced by a driver under the same conditions?
Answer:
Acceleration = 18g
Explanation:
Given the following data;
Initial velocity, u = 26m/s
Final velocity, v = 0
Time = 0.15 secs
To find the acceleration;
In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.
This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.
Hence, if we subtract the initial velocity from the final velocity and divide that by the time, we can calculate an object’s acceleration.
Mathematically, acceleration is given by the equation;
[tex]Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}[/tex]
Substituting into the equation, we have;
[tex]a = \frac{0 - 26}{0.15}[/tex]
[tex]a = \frac{26}{0.15}[/tex]
Acceleration = 173.33m/s2
To express it in magnitude of g;
Acceleration = 173.33/9.8
Acceleration = 17.7 ≈ 18g
Acceleration = 18g
Two kids are roller skating. Amy, with a mass of 55 kg, is traveling forward at 3 m/s. Jenny, who has a mass of 40 kg, is traveling in the opposite direction at 5 m/s. They crash into each other and hold onto each other so that they move as one mass. How fast are they traveling?
Answer:
-7/19
Explanation:
If Earth’s Moon were replaced with a typical neutron star, what would the angular diameter of the neutron star be as seen from Earth?
Answer:
[tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex]
Explanation:
d = Diameter of typical neutron star = 20 km = 20000 m
D = Distance between Earth and Moon = [tex]384.4\times 10^6\ \text{m}[/tex]
Here, [tex]D>>d[/tex] so we use small angle approximation
[tex]\delta=\dfrac{d}{D}\\\Rightarrow \delta=\dfrac{20000}{384.4\times 10^6}\\\Rightarrow \delta=0.00005202\ \text{rad}=\dfrac{0.00005202\times 180}{\pi}=0.003^{\circ}[/tex]
The angular diameter of the neutron star would be [tex]0.00005202\ \text{rad}=0.003^{\circ}[/tex] from Earth.
You are designing a flywheel. It is to start from rest and then rotate with a constant angular acceleration of 0.200 rev/s^2. The design specifications call for it to have a rotational kinetic energy of 330 J after it has turned through 30.0 revolutions.
What should be the moment of inertia of the flywheel about its rotation axis?
Express your answer with the appropriate units.
Answer: 1.14 kg*m/s
Explanation:
The first person explained everything right, they just forgot to convert the angular acceleration to rads/sec^2 from revs/sec^2. Once that is converted, your answer should come out right.
Another small thing, the answer there has an extra unnecessary step. It tells you to find the square root of w^2 to find w but that is unnecessary since the final equation asked for w^2. Hope this helps! :)
The moment of inertia I of the flywheel about its rotation axis is
[tex]1.39Kgm^2[/tex]Given
Angular displacement,
[tex]\theta = 30rev \\\\\theta = (30) * 2\pi rad \\\\\theta = 188.495rad[/tex]
Therefore, Final angular velocity (w) will be:
[tex]w^2 = 2\alpha\theta\\\\w^2 = 2 * (0.200 * 2\pi) * (188.49)\\\\w^2 = 473.73\\\\w = 21.76 rad/s[/tex]
Therefore,
moment of inertia
[tex]I = 2 * K / w^2[/tex]
[tex]I = 2 * 330 / 473.73[/tex]
[tex]I = 1.39kgm^2[/tex]
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a bus with a mass of 5000kg is uniformly acceleration from rest. The net force acting on the bus is
Answer:
50,000N
Explanation:
According to Newton's second law of motion;
Net Force = Mass * acceleration
Given
Mass = 5000kg
Let the acceleration = 10m/s²
Net force = 5000 * 10
Net force = 50,000N
Hence the net force acting on the bus is 50000N
A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.
(a) Neglecting air resistance, how long does it take the skier to reach the ground?
(b) How far horizontally does the skier travel in the air before landing?
m
Explanation:
Given
Velocity v = 23.0m/s
Distance S = 3.45m
Required
Time it will take the skier to reach the ground;
Using the equation of motion;
S = ut + 1/2gt²
3.45 = 23t + 1/2(9.8)t²
3.45 = 23t + 4.9t²
4.9t²+23t-3.45 = 0
Factorize;
t = -23 ±√23²-4(4.9)(-3.45)/2(4.9)
t = -23 ±√529+67.62/9.8
t = -23±√596.62/9.8
t = -23±24.43/9.8
t = 1.43/9.8
t = 0.146 secs
Hence take the skier 0.146 secs to reach the ground.
b) Horizontal distance covered is the range;
Range = U√2H/g
Range = 23√2(3.45)/9.8
Range = 23√6.9/9.8
Range = 23√0.7041
Range = 23(0.8391)
Range = 19.29m
Hence the horizontal distance travelled in air is 19.29m
(a).The time taken by the skier to reach the ground is 0.145 second.
(b).The skier travel in the air before landing is 19.29 meter.
a. Given that A skier leaves the end of a horizontal ski jump at 23.0 m/s and falls through a vertical distance of 3.45 m before landing.
Using equation of motion.
[tex]S=ut+\frac{1}{2}gt^{2}[/tex]
Where S is vertical distance , u is initial velocity and g is gravitational acceleration.
Substitute S = 3.45 m, u = 23m/s and g = 9.8 in above equation.
[tex]3.45=23t+\frac{1}{2} (9.8)t^{2}\\\\4.9x^{2} +23t-3.45=0\\\\t=0.145,-4.83[/tex]
Since, time can not be negative.
So that, [tex]t=0.145s[/tex]
b. The horizontal distance travel before landing is known as Range.
Horizontal distance ,
[tex]=v*\sqrt{\frac{2S}{g} }[/tex]
Substitute v = 23m/s , S = 3.45 and g = 9.8 meter per second square.
[tex]Distance=23*\sqrt{\frac{2*3.45}{9.8} } \\\\Distance=23*\sqrt{0.7041} \\\\Distance=23*0.8391=19.29m[/tex]
Thus, The skier travel in the air before landing is 19.29 meter.
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. A car going initially with a velocity 15 m/s accelerates at a rate of 2 m/s2 for 10 seconds. It then accelerates at a rate of -1.5 m/s until stop. Find the car’s maximum speed. Calculate the total distance traveled by the car.
Answer:
The maximum speed of the car is 35 m/s
The total distance traveled by the car is 658.33 m
Explanation:
Given;
initial velocity of the car, u = 15 m/s
acceleration of the car, a = 2 m/s²
time of car motion, t = 10 s
(i)
Initial distance traveled by the car is given by;
d₁ = ut + ¹/₂at²
d₁ = (15 x 10) + ¹/₂(2)(10)²
d₁ = 150 + 100
d₁ = 250 m
The maximum speed of the car during this is given by;
v² = u² + 2ad₁
v² = (15)² + (2 x 2 x 250)
v² = 1225
v = √1225
v = 35 m/s
(ii)
The final distance cover by the car during the deceleration of 1.5 m/s².
Note: the final or maximum speed of the car becomes the initial velocity during deceleration.
v² = u² + 2ad₂
where;
v is the final speed of the car when it stops = 0
0 = u² + 2ad₂
0 = (35²) + (2 x - 1.5 x d₂)
0 = 1225 - 3d₂
3d₂ = 1225
d₂ = 1225 / 3
d₂ = 408.33 m
The total distance traveled by the car is given by;
d = d₁ + d₂
d = 250 m + 408.33 m
d = 658.33 m
If a dog has a mass of 2.5 kg, what is its weight and what is the normal force that it feels.
I
Answer:
Weight = normal force = 24.5 N
Explanation:
Given that,
Mass of a dog, m = 2.5 kg
We need to find its weight and the normal force that it feels.
The weight of an object is given by :
W = mg
Where g is the acceleration due to gravity
[tex]W=2.5\times 9.8\\\\=24.5\ N[/tex]
The normal force is balanced by the weight of an object. So,
Weight = normal force = 24.5 N
A 5.3 kg block rests on a level surface. The coefficient of static friction is μ_s=0.67, and the coefficient of kinetic friction is μ_k= 0.48 A horizontal force, x is applied to the block. As x is increased, the block begins moving. Describe how the force of friction changes as x increases from the moment the block is at rest to when it begins moving. Show how you determined the force of friction at each of these times ― before the block starts moving, at the point it starts moving, and after it is moving. Show your work.
As the pushing force x increases, it would be opposed by the static frictional force. As x passes a certain threshold and overcomes the maximum static friction, the block will start moving and will require a smaller magnitude x to maintain opposition to the kinetic friction and keep the block moving at a constant speed. If x stays at the magnitude required to overcome static friction, the net force applied to the block will cause it to accelerate in the same direction.
Let w denote the weight of the block, n the magnitude of the normal force, x the magnitude of the pushing force, and f the magnitude of the frictional force.
The block is initially at rest, so the net force on the box in the horizontal and vertical directions is 0:
n + (-w) = 0
n = w = m g = (5.3 kg) (9.80 m/s²) = 51.94 N
The frictional force is proportional to the normal force, so that f = µ n where µ is the coefficient of static or kinetic friction. Before the block starts moving, the maximum static frictional force will be
f = 0.67 (51.94 N) ≈ 35 N
so for 0 < x < 35 N, the block remains at rest and 0 < f < 35 N as well.
The block starts moving as soon as x = 35 N, at which point f = 35 N.
At any point after the block starts moving, we have
f = 0.48 (51.94 N) ≈ 25 N
so that x = 25 N is the required force to keep the block moving at a constant speed.
As x is increasing it will be opposed by a static frictional force and for the object to start moving and maintain its acceleration, the magnitude of x must exceed the magnitude of the static frictional force and kinetic frictional force
Magnitude of normal force ( object at rest ); n = 51.94 N Required magnitude of x before the movement of object ; x = 35 NMagnitude of x after object start moving x = 25 NGiven data :
mass of block at rest ( m ) = 5.3 kg
Coefficient of static friction ( μ_s ) =0.67
Coefficient of kinetic friction is ( μ_k ) = 0.48
Horizontal force applied to block = x
First step : magnitude of normal force ( n ) when object is at rest
n = w where w = m*g
n - w = 0
n - ( 5.3 * 9.81 ) = 0 ∴ n = 51.94 N
Second step : Required magnitude of x before the movement of object
F = μ_s * n
F = 0.67 * 51.94 = 34.79 N ≈ 35 N
∴ The object will start moving once F and x = 35 N
Final step : Magnitude of x after object start moving
F = μ_k * n
= 0.48 * 51.94 = 24.93 N ≈ 25 N
∴ object will continue to accelerate at a constant speed once F and x = 25N
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A rolling ball moves from x1 = 8.0 cm to x2 = -4.1 cm during the time from t1 = 2.9 s to t2 = 6.0 s .
Complete Question
A rolling ball moves from [tex]x_1 = 8.0 \ cm[/tex] to [tex]x_2 = - 4.1 \ cm[/tex] during the time from [tex]t_1 = 2.9 s[/tex] to [tex]t_2 = 6.0s[/tex]
What is its average velocity over this time interval?
Answer:
The velocity is [tex]v = 3.903 \ m/s[/tex]
Explanation:
From the question we are told that
The first position of the ball is [tex]x_1 = 8.0 \ cm[/tex]
The second position of the ball is [tex]x_2 = - 4.1 \ cm[/tex]
Generally the average velocity is mathematically represented as
[tex]v = \frac{ x_1 - x_2}{t_2 - t_1}[/tex]
=> [tex]v = \frac{ 8 - -4.1 }{ 6 - 2.9 }[/tex]
=> [tex]v = 3.903 \ m/s[/tex]
a current of 200mA through a conductor converts 40 joules of electrical energy into heat in 30 seconds determine the potential drop across the conductor
Answer:
V = 6.65 [volt]
Explanation:
We must first find the power generated, power is defined as the amount of energy consumed or generated in a given time.
[tex]P=\frac{E}{t}[/tex]
where:
P = power [w]
E = energy = 40 [J]
t = time = 30 [s]
[tex]P =40/30\\P = 1.33[w][/tex]
Now we can calculate the voltage or potential drop by means of the power, the power is calculated by means of the product of the voltage by the current.
[tex]P =V*I[/tex]
where:
V = voltage [volts]
I = current = 200mA = 0.2 [A]
[tex]V = P/I\\V = 1.33/0.2\\V = 6.65 [Volt][/tex]
A submarine sends out a sonar signal (sound waves) in a direction directly downward it take 2.3 s for the sound waves to travel from the submarine to the ocean bottom and back to the submarine how high (approx) up from the ocean floor is the submarine?speed of sound in water is 1490 m/s
Explanation:
Using the formula;
2x = vt
x is the distance up from the ocean floor the submarine is
v is the speed of sound in water
t is the time
Given
t = 2.3s
v = 1490m/s
Required
how high (approx) up from the ocean floor is the submarine x
From the formula;
x = vt/2
x = 1490(2.3)/2
x = 745(2.3)
x = 1,713.5m
Hence the submarine is 1713.5m high up from the ocean floor
A spaceship is accelerating at 1000 m/sec2 . How much force is required from the backthrusters to completely stop the spaceship?
Answer:
1000x Newton
Explanation:
Step one
given data
acceleration= 1000 m/s²
The question did not specify the mass of the mass of the space ship.
So, let's assume the mass is x kg
Step two:
Required is the force F in Newton
From Newtons first law, it states that a body will continue to be at rest or uniform motion unless acted upon by a force.
F=mass x Acceleration
F=ma
Substituting our given data we have
F=1000x Newton
a 1000kg car uses a breaking force of 10,000N to stop in two second. What impulse acts on the car?
Answer:
5,000
Explanation:
Vf = Vi + a * t
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s.
Complete Question
A small object moves along the x-axis with acceleration ax(t) = −(0.0320m/s3)(15.0s−t)−(0.0320m/s3)(15.0s−t). At t = 0 the object is at x = -14.0 m and has velocity v0x = 7.10 m/s. What is the x-coordinate of the object when t = 10.0 s?
Answer:
The position of the object at t = 10s is [tex]X = 38.3 \ m[/tex]
Explanation:
From the question we are told that
The acceleration along the x axis is [tex]a_{x}t = -(0.0320\ m/s^3)(15.0 s- t)- (0.0320\ m/s^3)[/tex]
The position of the object at t = 0 is x = -14.0 m
The velocity at t = 0 s is [tex]v_{0}x = 7.10 m/s[/tex]
Generally from the equation for acceleration along x axis we have that
[tex]a_x = \frac{dV_{x}}{dt} = -0.032 (15- t)[/tex]
=> [tex]\int\limits {dV_{x}} \, = \int\limits {-0.032(15- t)} \, dt[/tex]
=> [tex]V_{x} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
At t =0 s and [tex]v_{0}x = 7.10 m/s[/tex]
=> [tex]7.10 = -0.032 [15(0) - \frac{(0)^2 }{2} ]+ K_1[/tex]
=> [tex]K_1 = 7.10[/tex]
So
[tex]\frac{dX}{dt} = -0.032 [15t - \frac{t^2 }{2} ]+ K_1[/tex]
=> [tex]\int\limits dX = \int\limits [-0.032 [15t - \frac{t^2 }{2} ]+ K_1] }{dt}[/tex]
=> [tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ K_1t +K_2[/tex]
At t =0 s and x = -14.0 m
[tex]-14 = -0.032 [ 15\frac{0^2}{2} - \frac{0^3 }{6} ]+ K_1(0) +K_2[/tex]
=> [tex]K_2 = -14[/tex]
So
[tex]X = -0.032 [ 15\frac{t^2}{2} - \frac{t^3 }{6} ]+ 7.10 t -14[/tex]
At t = 10.0 s
[tex]X = -0.032 [ 15\frac{10^2}{2} - \frac{10^3 }{6} ]+ 7.10 (10) -14[/tex]
=> [tex]X = 38.3 \ m[/tex]