A skier is accelerating down a theta = 33.8o hill at a =3.55m/s2, as seen in the

Question

A skier is accelerating down a theta = 33.8o hill at a =3.55m/s2, as seen in the figure below. a) What is the magnitude (size) of the vertical component of her acceleration? Enter units.

The velocity vector V1 has a magnitude of 5m/s and is directed along the +x-axis. The velocity vector V2 has a magnitude of 4m/s. The sum of the two is V3, so that V3 = V1+V2
(For each statement select T-True, or F-False; If the first is True and the rest F, enter TFFFFF).

A) The magnitude of V3 can be 9m/s
B) The x-component of V3 can be 0m/s
C) The magnitude of V3 can be -7m/s
D) The magnitude of V3 can be 0
E) The magnitude of V3 can be 10m/s
F) The magnitude of V3 can be 8m/s

Explanation / Answer

1) Let Ay = vertical component of acceleration
Ay = (3.55 m/s^2)(sin33.8) = 1.97 m/s^2

2) A)  True , if V2 i directed to + x-axis

B) False, as V1 is directed along the +x-axis.

C)   False. Magnitudes are never negative.

D) False, V3 min = 5 - 4 = 1 m/s

E)  False. If V1 and V2 are aligned, the maximum sum they can have is 9 m/s

F) True. Arrange V1 and V2 such that the angle between them is about 54.9 [deg] and V3's magnitude is indeed 8 m/s.

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