Problem Eight. A snowmobile is originally at the point with position vector 31.1 m at 95.5° counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant acceleration 1.73 m/s² at 200°. After 5.00 s have elapsed, find the following. 9.) The velocity vector in m/s. (A)=-4.38+0.185ĵ (D) = 0.185 +4.38ĵ (B)=0.1851-4.38ĵ (E) = 4.38 +0.185ĵ (C) v=-0.1851-4.38ĵ (A)=-39.3-4.30ĵ 10.) The final position vector in meters. (B)=39.3-4.30ĵ (C) = -4.61 +39.3ĵ (D) = 39.31 +4.30ĵ (E) = 4.30 +39.3ĵ
Problem Eight. A snowmobile is originally at the point with position vector 31.1 m at 95.5° counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant acceleration 1.73 m/s² at 200°. After 5.00 s have elapsed, find the following. 9.) The velocity vector in m/s. (A)=-4.38+0.185ĵ (D) = 0.185 +4.38ĵ (B)=0.1851-4.38ĵ (E) = 4.38 +0.185ĵ (C) v=-0.1851-4.38ĵ (A)=-39.3-4.30ĵ 10.) The final position vector in meters. (B)=39.3-4.30ĵ (C) = -4.61 +39.3ĵ (D) = 39.31 +4.30ĵ (E) = 4.30 +39.3ĵ
University Physics Volume 1
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Chapter2: Vectors
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Problem 29P: A delivery man starts at the post office, chives 40 km north, then 20 km west, then 60 km northeast,...
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
Transcribed Image Text:Problem Eight. A snowmobile is originally at the point with position vector 31.1 m at 95.5°
counterclockwise from the x-axis, moving with velocity 4.89 m/s at 40.0°. It moves with constant
acceleration 1.73 m/s² at 200°. After 5.00 s have elapsed, find the following.
9.) The velocity vector in m/s.
(A)=-4.38+0.185ĵ
(D) = 0.185 +4.38ĵ
(B)=0.1851-4.38ĵ
(E) = 4.38 +0.185ĵ
(C) v=-0.1851-4.38ĵ
(A)=-39.3-4.30ĵ
10.) The final position vector in meters.
(B)=39.3-4.30ĵ
(C) = -4.61 +39.3ĵ
(D) = 39.31 +4.30ĵ
(E) = 4.30 +39.3ĵ
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