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