A point has a position vector given by r of t equals the vector with components t, 4 , for all time t ≥ 0 seconds. Find the speed of the object at t = 3 seconds.

Question

A point has a position vector given by  , for all time t ≥ 0 seconds. Find the speed of the object at t = 3 seconds

A
B	1
C	0
D

Answer 1

r(t) = (t2+1, 4-t)

dr/dt = r(t) = (2t, -1)

|| r(t) || = sqrt (4t² +1)

speed of object at t=1 sec

|| dr/dt|| = sqrt (4² +1² )

= sqrt 5

Answer 2

The position vector equation provides the position of the particle in two coordinates, let's say x and y. So if r(t) = (t, 4), the two coordinates are (3, 4) at t = 3 seconds.

However, we are interested in the speed of the object, so we should take the derivative of each component of the position vector. This means that v(t) = r'(t) = (1, 4). So the particle is moving steadily with a velocity of 1 in the x direction and 4 in the y direction.

Use the Pythagorean theorem to find the magnitude of the velocity with these two components:

v = √(v_x² + v_y²) = √(1² + 4²) = √5

The answer is D.