Please show all work! The position of an object moving along an x axis is given

Question

Please show all work!

The position of an object moving along an x axis is given by x(t)= -5t^3 + 3t^2 + t, where x is expressed in meters and t is given in seconds, Determine the position, velocity and acceleration of that object at t = 2s. What is the maximum positive velocity reached by the object and at what time is it reached? Determine the average velocity of the object under consideration between t = 1s and t = 2s. A dart is thrown horizontally with an initial speed of 11 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.28 s later. Neglect air resistance. What is distance PQ? How far away from the dart board is the dart released? Two vectors are presented in unit-vector notation as A vector = -4 i cap + 2kcap and B vector = 3i cap - k cap. Find A vector times B vector. A vector middot B vector, the component of vector B vector along the direction of vector A vector and the angle between vectors A vector and B vector. The figure below shows an overhead view of a 50 g disk and two of the three horizontal forces that act on it as it is on a frictionless table. The force magnitudes are F_1 = 6N and F_2 = 7N, while the indicated angles are theta_1 = 30 degree and theta_2 = 45 degree. In the unit-vector notation, what is the third force if the disk has constant acceleration a vector = (13 i cap -14 j cap) m/s^2, where t is time? Using only one sentence, define the subject of physics. What is kinematics? What is dynamics? Newton's first law. Newton's second law. Newton's third law.

Explanation / Answer

1) given x(t) = -5*t^3 + 3*t^2 + t

a) at t = 2s,

position, x(t) = -5*2^3 + 3*2^2 + 2

= -26 m

velocity, v(t) = dx(t)/dt

= -5*3*t^2 + 3*2*t

at t = 2s

v(t) = -5*3*2^2 + 3*2*2

= -48 m/s

acceleration, a(t) = dv(t)/dt

= -15*2*t + 6

at t = 2s

a(t) = -15*2*2 + 6

= -54 m/s^2

b) when v is maximum, dvmax/dt = 0

-30*t + 6 = 0

t = 6/30

= 1/5 s

so, vmax = -5*3*(1/5)^2 + 3*2*(1/2)

= 2.4 m/s

c) x(1) = -5*1^3 + 3*1^2 + 1 = -1 m

x(2) = -5*2^3 + 3*2^2 + 2 = -26 m

Vavg = ( x(2) - x(1))/(t2 - t1)

= (-26 - (-1))/(2-1)

= -25 m/s

1) given x(t) = -5*t^3 + 3*t^2 + t

a) at t = 2s,

position, x(t) = -5*2^3 + 3*2^2 + 2

= -26 m

velocity, v(t) = dx(t)/dt

= -5*3*t^2 + 3*2*t

at t = 2s

v(t) = -5*3*2^2 + 3*2*2

= -48 m/s

acceleration, a(t) = dv(t)/dt

= -15*2*t + 6

at t = 2s

a(t) = -15*2*2 + 6

= -54 m/s^2

b) when v is maximum, dvmax/dt = 0

-30*t + 6 = 0

t = 6/30

= 1/5 s

so, vmax = -5*3*(1/5)^2 + 3*2*(1/2)

= 2.4 m/s

c) x(1) = -5*1^3 + 3*1^2 + 1 = -1 m

x(2) = -5*2^3 + 3*2^2 + 2 = -26 m

Vavg = ( x(2) - x(1))/(t2 - t1)

= (-26 - (-1))/(2-1)

= -25 m/s

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