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.
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
To determine the correct answer, let's evaluate the given velocity vector at t = 1.
The velocity vector for the object is (√(36-6t), 4t). Substituting t = 1 into the expression, we get (√(36-6(1)), 4(1)) = (√30, 4).
Now let's analyze each of the answer choices:
A. The object is at rest: If an object is at rest, its velocity vector should be (0, 0). Since the given velocity vector at t = 1 is (√30, 4), this statement is not true.
B. The speed of the object is √30 + 4: The speed of an object is the magnitude of its velocity vector. The magnitude of the given velocity vector at t = 1 is ||(√30, 4)|| = √(30 + 16) = √46. Therefore, this statement is not true.
C. The object is speeding up: To determine if the object is speeding up or slowing down, we can evaluate the acceleration. The acceleration vector is the derivative of the velocity vector. Taking the derivative of (√(36-6t), 4t) with respect to t, we get (√6, 4).
Since the acceleration vector (√6, 4) is constant and does not change with t, the object is neither speeding up nor slowing down. Therefore, this statement is not true.
D. The object is slowing down: Similar to the explanation for choice C, the acceleration vector (√6, 4) is constant and does not change with t. Therefore, the object is not slowing down. Hence, this statement is not true.
Given that none of the statements are true, the correct answer is actually None of the above.