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

Question

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

Question

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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Physics Khải Quang 2 years 2021-07-27T16:18:01+00:00 2021-07-27T16:18:01+00:00 1 Answers 5 views 0

Answers ( )

thnahdat · Answer 1 · 2021-07-27T16:19:45+00:00

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)