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2 years ago

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Physics

1 answer:

alexgriva [62]2 years ago

6 0

Answer:

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Explanation:

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The velocity function (in meters per second) is given for a particle moving along a line. v(t) = 5t − 9, 0 ≤ t ≤ 3 (a) Find the

Reil [10]

Answer:

a) The displacement is -4.5 m.

b) The traveled distance is 11.7 m.

Explanation:

Hi there!

a)The velocity of the particle is the derivative of the displacement function, x(t):

v(t) = dx/dt = 5t - 9

Separating varibles:

dx = (5t - 9)dt

Integrating both sides from x = x0 to x and from t = 0 to t.

x - x0 = 1/2 · 5t² - 9t

x = 1/2 · 5t² - 9t + x0

If we place the origin of the system of reference at x = x0, the displacement at t = 3 will be x(3):

x(3) = 1/2 · 5 · (3)² - 9(3) + 0

x(3) = -4.5

The displacement at t = 3 s is -4.5 m. It means that the particle is located 4.5 m to the left from the origin of the system of reference at t = 3 s.

b) When the velocity is negative, the particle moves to the left. Let´s find the time at which the velocity is less than zero:

v = 5t - 9

0 > 5t - 9

9/5 > t

1.8 s > t

Then until t = 1.8 s, the particle moves to the left from the origin of the reference system.

Let´s find the position of the particle at that time:

x = 1/2 · 5t² - 9t

x = 1/2 · 5(1.8 s)² - 9(1.8 s)

x = -8.1 m

From t = 0 to t = 1.8 s the traveled distance is 8.1 m. After 1.8 s, the particle has positive velocity. It means that the particle is moving to the right, towards the origin. If at t = 3 the position of the particle is -4.5 m, then the traveled distance from x = -8.1 m to x = -4.5 m is (8.1 m - 4.5 m) 3.6 m.

Then, the total traveled distance is (8.1 m + 3.6 m) 11.7 m.

4 0

2 years ago

An airplane is flying 340 km/hr at 12o east of north. the wind is blowing 40 km/hr at 34o south of east. what is the plane's act

jok3333 [9.3K]

Define an x-y coordinate system such that
The positive x-axis = the eastern direction, with unit vector $\hat{i}$ .
The positive y-axis = the northern direction, with unit vector $\hat{j}$ .

The airplane flies at 340 km/h at 12° east of north. Its velocity vector is
$\vec{v}_{1} = 340(sin(15^{o})\hat{i} + cos(15^{o})\hat{j} ) = 88\hat{i} + 328.4\hat{j}$

The wind blows at 40 km/h in the direction 34° south of east. Its velocity vector is
$\vec{v}_{2} =40(cos(34^{o})\hat{i} - sin(24^{o})]\hat{j}) = 33.1615\hat{i} -22.3677\hat{j})$

The plane's actual velocity is the vector sum of the two velocities. It is
$\vec{v}=\vec{v}_{1}+\vec{v}_{2} = 121.1615\hat{i}+306.0473\hat{j}$

The magnitude of the actual velocity is
v = √(121.1615² + 306.0473²) = 329.158 km/h

The angle that the velocity makes north of east is
tan⁻¹ (306.04733/121.1615) = 21.6°

Answer:
The actual velocity is 329.2 km/h at 21.6° north of east.

5 0

2 years ago

The main source of information used by astronomers to learn about objects in space is.

Dmitriy789 [7]

Electromagnetic waves

4 0

1 year ago

The wheels of a car have radius 12 in. and are rotating at 600 rpm. Find the speed of the car in mi/h.

guajiro [1.7K]

Answer:

21.4 mph

Explanation:

Circumference of tire in FEET = pi * d = pi * 1 ft = pi feet

pi feet x 600 rot/min * 60 min /hr * 1 mile / 5280 feet = 21.4 mph

7 0

6 months ago

In which part of the scientific method do you tell why the research is being done?

wel

You tell it in pupouse

5 0

3 years ago

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