Asked by Anonymous
An object moves in the xy-plane so that its velocity vector for 0 <= t <= 10 is (√(36-6t), 4t). Which of these statements is true about the object at t = 1?
A. The object is at rest.
B. The speed of the object is √30 + 4.
C. The object is speeding up.
D. The object is slowing down.
My answer is C.
A. The object is at rest.
B. The speed of the object is √30 + 4.
C. The object is speeding up.
D. The object is slowing down.
My answer is C.
Answers
Answered by
Damon
Vx = (36 - 6t)^.5 at t = 1, it is sqrt(30)
Vy = 4t at t = 1, it is 4
so |V| at 1 = sqrt(46)
Ax = .5 (36-6t)^-.5 * (-6) = -3/sqrt 30 = -.55 at t = 1
Ay = 4
The velocity is increasing fast, yes, C
Vy = 4t at t = 1, it is 4
so |V| at 1 = sqrt(46)
Ax = .5 (36-6t)^-.5 * (-6) = -3/sqrt 30 = -.55 at t = 1
Ay = 4
The velocity is increasing fast, yes, C
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.