INTERPRETING TANGENTS If an arrow is shot upward on the moon with a velocity of

Question

INTERPRETING TANGENTS If an arrow is shot upward on the moon with a velocity of 58 meters per second, its height (meters) after t seconds can be found by h(t) = 58 t - 0.83 t2. Each graph shows h(t) along with a tangent to the curve of h(t) at a given point. (1) What do the x- and y-coordinates of (4) What do the x- and y-coordinates of (7) What do the x- and y-coordinates of the point (1, 57.17) represent? the point (34.93975894737, 1013.25) represent? the point (69.879518072289, 0) represent? Describe the meaning of each coordinate Describe the meaning of each coordinate Describe the meaning of each coordinate in the context of the problem. in the context of the problem. in the context of the problem. (2) Determine the equation of the tangent (5) Determine the equation of the tangent (6) Determine the equation of the tangent line to the curve of h(t) at line to the curve of h(t) at line to the curve of h(t) at (1, 57.17). (34.93975894737, 1013.25). (69.879518072289, 0). Note: the tangent has a positive slope. Note: the tangent is horizontal. Note: the tangent has a negative slope. (3) Describe the motion of the arrow (6) Describe the motion of the arrow (9) Describe the motion of the arrow in terms of direction and speed (i.e., velocity) in terms of direction and speed (i.e., velocity) in terms of direction and speed at 1 second into the problem. at 34.93975894737 seconds. at 69.879518072289 seconds.

Explanation / Answer

(a) velocity is given by dH/dt = V(t) = 58 - 1.66t m/sec
So that V(a) = 58 - 1.66a

(b) After one second, a = 1; therefore V(1) = 58 - 1.66
So V(1) = 56.33 m/sec

(c) The arrow will hit the moon when it has travelled up and back down, and so H=0
Therefore 58t - 0.83t^2 = 0
There are two solutions:
t=0 (not started yet)
t=58/0.83 = 69.88 sec (gone up and back down)

(d) At t = 69.88, V = 58 - 1.66x69.88 = -58 m/sec
In other words, it has returned to the ground with the same velocity it had when launched but travelling in the opposite direction.

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