Question

5.00-kg particle starts from the origin at time zero. Its velocity as a function of time is given by v =6t^2 i + 2t j where v is in meters per second and t is in seconds. (a) Find its position as a function of time. (b) Describe its motion qualitatively. Find (c) its acceleration as a function of time, (d) the net force exerted on the particle as a function of time, (e) the net torque about the origin exerted on the particle as a function of time, (f) the angular momentum of the particle as a function of time, (g) the kinetic energy of the particle as a function of time, and (h) the power injected into the system of the particle as a function of time.

Answer 1

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 = $\frac{dr}{dt}$  

          dr = v dt

we substitute and integrate.

        ∫ dr = ∫ (6 t² i + 2t j) dt

        r - 0 = $6 \frac{t^3 }{3} \ \hat i + 2 \frac{t^2}{2 \ \hat j }$  

       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.

           $a = \frac{dv}{dt}$  

           a = $\frac{d}{dt} \ (6t^2 i + 2t j)$  

           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.

         $\tau = \left[\begin{array}{ccc}i&j&k\\F_x&F_y&F_z\\x&y&z\end{array}\right]$  

        $\tau = \left[\begin{array}{ccc}i&j&k\\60t &10&0\\2t^3 &t^2&0\end{array}\right]$  

       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.

       $\left[\begin{array}{ccc}i&j&k\\2t^3 &t^2&0\\6t^2&2t&0\end{array}\right]$  

    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 = $\frac{dW}{dt}$dW / dt

The relationship between work and kinetic energy

           W = ΔK

     

          P = $2m \ \frac{d}{dt} ( 9 t^4 + t^2)$

          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

Answer 2

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. ...

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