The position of a particle moving in an xy plane is given by with in meters and t in seconds. In unit-vector notation, calculate (a), (b), and (c) for t = 2.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t = 2.00 s? Give your answer in the range of (-180o; 180o)

a) r(@2sec) = (6.00i - 106j) m ; b) v = ( 19i - 224t³j) m/sec ; c) a = (24i - 336j) m/sec² ; d) θ = -85.1516°

Explanation:

Correct question statement:

The position r of a particle moving in an xy plane is given by r = (2.00t³ - 5.00t)° + (6.00 -7.00t 4 ) ĵ , with r in meters and t in seconds. In unit-vector notation, calculate (a) r , (b) v , find (c) a for t = 2.00 s. (d) What is the angle between the positive direction of the x axis and a line tangent to the particle's path at t= 2.00 s?

Given Data:

r = (2t³ - 5t)i + ( 6 - 7t⁴)j equation (1)

a)

r(@2sec) = (2(2)³ - 5(2))i + ( 6 - 7(2)⁴)j

               = (6.00i - 106j) m

b)

By taking derivative of equation 1 once, we get

              v(t) = ( 6.00t²-5.00 )i - 28t³j equation (2)

By putting t = 2sec, we get

                 v = ( 19i - 224t³j) m/sec

c)

By differentiating equation 1 twice of equation 2 once, we get

              a(t) = 12ti - 84t²j

At t = 2 sec, we get

                 a = (24i - 336j) m/sec²

d)

θ = tan ⁻¹ (-224/19) = -85.1516° or 94.8483°

   θ = -85.1516° it is also equivalent to 274.8484°(counterclockwise from +x-axis), and it is in 4th quadrant.

hi there!

your question asks at which segment does freezing occur.

since we're talking about water, the freezing point of water will be 32 degrees fahrenheit or below.

hope this ! ~

ice to water 1 water to vapour 2 latent heat