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The acceleration vector of a particle Q is given by the following acceleration vector a = 4i – 6tJ + sin(.2t)k m/s^2, where t is the time. A
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The acceleration vector of a particle Q is given by the following acceleration vector a = 4i – 6tJ + sin(.2t)k m/s^2, where t is the time. At t=0, the pointe Q is located at (x0,y0,z0) = (1,3,-5)m, and has a velocity vo= 2i -7j +8.4k m/s. Find the speed of Q and its distance from the starting point at t= 3s.
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2021-07-27T16:18:01+00:00
2021-07-27T16:18:01+00:00 1 Answers
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Answer:
V(3) =14i^ -34j^ +8.57 k^
S(3) =(25,-45,3.97)
Explanation:
We know that
V =a dt
from t=0 to 3s
V = 4i – 6tJ + sin(.2t)k m/s² dt
V =4t i^ – 3t^2j^ – cos(2t)/2 k^ +C
So we have
V(0) =-1/2 k^ +C =2i -7j +8.4 k
C=2i -7j +8.9k^
V =4t i^ – 3t^2j^ – cos(2t)/2 k^ + 2i -7j +8.9k^
Then putting 3s
V(3) =14i^ -34j^ +8.57 k^
Also
S(t) =V(t) dt
S(t) = 4t i^ – 3t^2j^ – cos(2t)/2 k^ + 2i -7j +8.9k^] dt
S(t)= (2t^2 +2t)i^ – (t^3 +7t) j^ -[sin(2t)/4 – 8.9t] k^ +C =i+3j-5k
So at t= 0 we have
S(0) = C= i+3j-5k
So S(t) =(2t^2 +2t)i^ – (t^3 +7t) j^ -[sin(2t)/4 – 8.9t] k^ +i+3j-5k
When t= 3
S(3) =25i -45j + 3.97k
S(3) =(25,-45,3.97)