this should read:If an arrow is shot upward on the moon with a velocity of 58 me
ID: 3082105 • Letter: T
Question
this should read: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 the point (1, 57.17) represent? Describe the meaning of each coordinate in the context of the problem. (4) What do the x- and y-coordinates of the point (34.93975894737, 1013.25) represent? Describe the meaning of each coordinate in the context of the problem. (7) What do the x- and y-coordinates of the point (69.879518072289, 0) represent? Describe the meaning of each coordinate in the context of the problem. (2) Determine the equation of the tangent line to the curve of h(t) at (1, 57.17). Note: the tangent has a positive slope. (5) Determine the equation of the tangent line to the curve of h(t) at (34.93975894737, 1013.25). Note: the tangent is horizontal. (6) Determine the equation of the tangent line to the curve of h(t) at (69.879518072289, 0). Note: the tangent has a negative slope. (3) Describe the motion of the arrow in terms of direction and speed (i.e., velocity) at 1 second into the problem. (6) Describe the motion of the arrow in terms of direction and speed (i.e., velocity) at 34.93975894737 seconds. (9) Describe the motion of the arrow in terms of direction and speed 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. As a matter of interest, the figure of 0.83 in your original equation is a little out. The average surface gravitational acceleration on the moon is 1.62 m/s^2, which would give 0.81 in the equation. The effect is small but it's there. Also, there are areas of slightly stronger gravitational attraction on the surface of the moon, located over the major craters. This is explained by the presence of heavier metal impact bodies under the surface, it is assumed.
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