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Answer:A 5.00 kg particle starts from the origin at the time zero. its velocity as function of time is v=6t^2i+2tj where v in meters per second and t ...
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(a) The position as function of time is x(t)=2t3i^+t2j^.x(t)=2t^3\hat\textbf i+t^2\hat\textbf j.x(t)=2t3i^+t2j^. (b) Describe the motion qualitatively. ...
Explanation:
The concept of derivatives and integrals allows to find the results for the questions are the motion of a particle where the speed depends on time are:
a)the position is: r = 2 t³ i + t² j
b) the position of the body on the y-axis is a parabola and on the x-axis it is a cubic function
c) The acceleration is: a = 12 t i + 2 j
d) the force is: F = 60 t i + 10 j
e) the torque is: τ = 40 t³ k^
f) tha angular momentum is: L = 4t³ - 6 t² k^
g) The kinetic energy is: K = 2 m t² (9t² +1)
h) The power is: P = 2m (36 t³ + 2t)
Kinematics studies the movement of bodies, looking for relationships between position, speed and acceleration.
a) They indicate the function of speed.
v = 6 t² i + 2t j
Ask the function of the position. The velocity is defined by the variation of the position with respect to time
v = [tex]\frac{dr}{dt}[/tex]
dr = v dt
we substitute and integrate.
∫ dr = ∫ (6 t² i + 2t j) dt
r - 0 = [tex]6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }[/tex]
r = 2 t³ i + t² j
b) The position of the body on the y axis is a parabola and on the x axis it is a cubic function.
c) Acceleration is defined as the change in velocity with time.
[tex]a = \frac{dv}{dt}[/tex]
a = [tex]\frac{d}{dt} \ (6t^2 i + 2t j)[/tex]
a = 12 t i + 2 j
d) Newton's second law states that force is proportional to mass times the body's acceleration.
F = ma
F = m (12 t i + 2 j)
F = 5 12 t i + 2 j
F = 60 t i + 10 j
e) Torque is the vector product of the force and the distance to the origin.
τ = F x r
The easiest way to write these expressions is to solve for the determinant.
[tex]\tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right][/tex]
[tex]\tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right][/tex]
τ = (60t t² - 2t³ 10) k
τ = 40 t³ k ^
f) Angular momentum
L = r x p
L =rx (mv)
L = m (rxv)
The easiest way to write these expressions is to solve for the determinant.
[tex]\left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right][/tex]
L = (4t³ - 6 t²) k
g) The kinetic energy is
K = ½ m v²
K = ½ m (6 t² i + 2t j) ²
K = m 18 t⁴ + 2t²
K = 2 m t² (9t² +1)
h) Power is work per unit time
P = [tex]\frac{dW}{dt}[/tex]dW / dt
The relationship between work and kinetic energy
W = ΔK
P = [tex]2m \ \frac{d}{dt} ( 9 t^4 + t^2)[/tex]
p = 2m (36 t³ + 2t)
In conclusion with the concept of derivatives and integrals we can find the results for the questions are the motion of a particle where the speed depends on time are:
a) The position is: r = 2 t³ i + t² j
b) The position of the body on the y-axis is a parabola and on the x-axis it is a cubic function
c) The acceleration is: a = 12 t i + 2 j
d) The force is: F = 60 t i + 10 j
e) The torque is: τ = 40 t³ k^
f) The angular momentum is: L = 4t³ - 6 t² k^
g) The kinetic energy is: K = 2 m t² (9t² +1)
h) The power is: P = 2m (36 t³ + 2t)
Learn more here: brainly.com/question/11298125
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