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Shkiper50 [21]
3 years ago
14

(a) Find the position vector of a particle that has the given acceleration and the specified initial velocity and position. a(t)

= 20t i + sin(t) j + cos(2t) k, v(0) = i, r(0) = j
Physics
1 answer:
dolphi86 [110]3 years ago
8 0

Answer:

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

Explanation:

a(t)=20t i+sin(t) j +cos(2t) k

v(t)=\int\limits^a_b {a(t)} \, dt

=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k---------------eqn 1

given v(0)=i

i=c_1i+(-1+c_2)j+(0+c_3)k

c_1=1   c_2=1  c_3=0

from equation 1

V(t)=(10t^2+c_1)i+(-cost+c_2)j+\frac{sin2t}{2}k----------eqn 2

now r(t)=\int\limits^a_b {v (t)} \, dt

(\frac{10t^3}{3}+t+c_1)i+(-sint+t+c_2)j+(\frac{-cos2t}{4}+c_3)k

given r(0)=j

0i+1j+ok = c_1i+c_2j+(\frac{-1}{4}+c_3)k

c_1=0  c_2=0  c_3 =  \frac{1}{4}

r(t)=(\frac{10t^3}{3}+t)i+(-sint+t+1)j+(\frac{-cos2t}{4}+\frac{1}{4})k

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Answer:

Explanation:

Given

Cross-sectional area of two areas is

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