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sweet-ann [11.9K]
3 years ago
10

) Suppose a particle travels along a straight line with velocity v(t) = t 2 e −3t meters per second after t seconds. How far doe

s the particle travel during the first 3 seconds? Round your answer to the nearest hundredth of a meter

Physics
2 answers:
pshichka [43]3 years ago
7 0

Answer:

x(t=3s) = 0.07 m to the nearest hundredth

Explanation:

v(t) = t² e⁻³ᵗ

Find displacement after t = 3 s.

Recall, velocity, v = (dx/dt)

v = (dx/dt) = t² e⁻³ᵗ

dx = t² e⁻³ᵗ dt

∫ dx = ∫ t² e⁻³ᵗ dt

This integration will be done using the integration by parts method.

Integration by parts is done this way...

∫ u dv = uv - ∫ v du

Comparing ∫ t² e⁻³ᵗ dt to ∫ u dv

u = t²

∫ dv = ∫ e⁻³ᵗ dt

u = t²

(du/dt) = 2t

du = 2t dt

∫ dv = ∫ e⁻³ᵗ dt

v = (-e⁻³ᵗ/3)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ (-e⁻³ᵗ/3) 2t dt

= (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt

But the integral (∫ 2t (-e⁻³ᵗ/3) dt) is another integration by parts problem.

∫ u dv = uv - ∫ v du

u = 2t

∫ dv = ∫ (-e⁻³ᵗ/3) dt

u = 2t

(du/dt) = 2

du = 2 dt

∫ dv = ∫ (-e⁻³ᵗ/3) dt

v = (e⁻³ᵗ/9)

∫ u dv = uv - ∫ v du

Substituting the variables for u, v, du and dv

∫ 2t (-e⁻³ᵗ/3) dt = 2t (e⁻³ᵗ/9) - ∫ 2 (e⁻³ᵗ/9) dt = 2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)

Putting this back into the main integration by parts equation

∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - ∫ 2t (-e⁻³ᵗ/3) dt = (-t²e⁻³ᵗ/3) - [2t (e⁻³ᵗ/9) + (2e⁻³ᵗ/27)]

x(t) = ∫ t² e⁻³ᵗ dt = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k (k = constant of integration)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + k

At t = 0 s, v(0) = 0, hence, x(0) = 0

0 = 0 - 0 - (2/27) + k

k = (2/27)

x(t) = (-t²e⁻³ᵗ/3) - 2t (e⁻³ᵗ/9) - (2e⁻³ᵗ/27) + (2/27)

At t = 3 s

x(3) = (-9e⁻⁹/3) - (6e⁻⁹/9) - (2e⁻⁹/27) + (2/27)

x(3) = -0.0003702294 - 0.0000822732 - 0.0000091415 + 0.0740740741 = 0.07361243 m = 0.07 m to the nearest hundredth.

harina [27]3 years ago
5 0

Answer:

-4.62×10‐4m

Explanation:

The velocity function of a body is simply the time derivative of its position x(t). This is obtained by usually differentiating the position function x(t) with respect to time t. That is,

V(t) = dx(t)/dt

In order to now derive x(t) from V(t) we have to integrate V(t) with respect to t. So

x(t) = ∫V(t)dt

In order to solve the above equation the knowledge of integration by parts would be required. For ease of writing the solution, I will present the steps to the answer in the attachment which can be found below.

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timurjin [86]

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initial velocity of the sports car, u = 0 m/s

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time of motion, t = 10 s

The kinetic energy of the car is given by;

K.E = ¹/₂m(v² - u²)

K.E = ¹/₂mv²

K.E = ¹/₂ x 1100 x 32²

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The average power of the engine of the sports car is given by;

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8 0
3 years ago
The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s
just olya [345]

Answer:

(a) 5610 kgm/s

(b) 5610 Ns.

(c)  78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S.I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv......................... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330(17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S.I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM................ Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S.I unit of force is Newton (N).

F = ma ................................. Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = ( v₂-v₁)/t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6)/60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330(0.2383)

F = 78.639 N.

F ≈ 78. 64 N

8 0
3 years ago
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