Please include full explanation. thank you Problem 3.40 Ite Cislants Part A A fa
ID: 1874339 • Letter: P
Question
Please include full explanation. thank you
Problem 3.40 Ite Cislants Part A A faulty model rocket moves in the n-lanc (thc positive-oroction is vertically upward) The rockets acceleration has components 14(1) = and adt) _ ?t, where 2 50 m/s1.8-9.00 m/s. and = 1 40 m/s3At 0 the rocket is at the ongin and has velocity ere-4x i + toy] with U0c- 1.00 m/s and vy 7.00m/s. Calculate the velocity vector as a function of time. Express your answer in terms of the t ly y, and a. Write the vector v(t) in the form v(t)o v t) where the x and y components are separated by comma Subrriit Incorrect. Try Again; 6 attempts remaining Part B Calculate the position vector as a tuncton of time Express your answer in terms of x, Uhy. B. T, and a. Write the vector rit) in the formt) rit)y where the x and y components are separated by a comma. SubmitExplanation / Answer
vector a(t) = [2.50t² i + (9.00 - 1.40t) j] m/s² for t in seconds
velocity:
v(t) = a(t) dt = [((/3)t³ + v0x) i + (ßt - t²/2 + v0y) j] m/s for t in seconds
v(t) = (v(t)x, v(t)y) = (t³/3 + v0x, ßt - t²/2 + v0y)
In case you need the numeric answer: v = [(2.50/3)t³ + 1.00) i + (9.00t - 0.70t² + 7.00) j] m/s
position:
r(t) = v(t) dt = [((/12)t + v0x*t) i + (ßt²/2 - t³/6 + v0y*t) j] m for t in seconds
r(t) = (t/12 + v0x*t, ßt²/2 - t³/6 + v0y*t)
In case you need the numeric answer: r(t) = (2.50t/12 + 1.00t, 4.50t² - 1.40t³/6 + 7.00t)
max height:
The maximum height is reached when v(t)y = 0:
9.00t - 0.70t² + 7.00 = 0
is quadratic with a positive root at t = 13.6 s. Then
r(t)y = 4.50*13.6² - 1.40*13.6³/6 + 7.00*13.6 = 341 m
distance at impact:
r(t)y = 0 = 4.50t² - 1.40t³/6 + 7.00t divide by t
0 = 4.50t - 1.40t²/6 + 7.00
is quadratic with a positive root at t = 20.7 s. So
r(20.7)x = 2.50*20.7/12 + 1.00*20.7 = 38 500 m