Your Mark: roblem # : Suppose that a particle has the following acceleration vec

Question

Your Mark: roblem # : Suppose that a particle has the following acceleration vector and initial velocity and position vectors. (a) Find the velocity of the particle at time t. (b) Find the position of the particle at time t Enter your answer as a symbolic function of t, as in these examples Enter the components of the velocity vector, separated with a comma. Problem #6(a): Enter your answer as a symbolic function of t, as in these examples Enter the components of the position vector, separated with a comma. Problem #6(b):

Explanation / Answer

6) We have given a(t)=10i+8tk,v(0)=7i-j,r(0)=j+6k

a) we know the velocity vector of a particle is v(t)=integration of (a(t))dt

v(t)=integration of (10i+8tk)dt

=10ti+(8t2k)/2 +C

v(t)=10ti+4t2k+C

plug v(0)=7i-j into v(t)

v(0)=10(0)i+4(0)2k+C=7i-j

0+0+C=7i-j implies C=7i-j

plug C=7i-j into v(t)

v(t)=10ti+4t2k+7i-j

=(10t+7)i+4t2k-j

v(t)=(10t+7)i+4t2k-j

the velocity of the particle at time t is v(t)=(10t+7)i-j+4t2k

b) we know the position vector of a particle is r(t)=integration of (v(t))dt

r(t)=integration of ((10t+7)i+4t2k-j)dt

=(10t2/2+7t)i+4t3k/3-tj+C

r(t)=(5t2+7t)i+(4/3)t3k-tj+C

plug r(0)=j+6k into r(t)

r(0)=(5(0)2+7*0)i+(4/3)(0)3k-0*j+C=j+6k

C=j+6k

plug C=j+6k into r(t)

r(t)=(5t2+7t)i+(4/3)t3k-tj+j+6k

=(5t2+7t)i+((4/3)t3+6)k+(1-t)j

r(t)=(5t2+7t)i+(1-t)j+((4/3)t3+6)k

the position of the particle at time t is r(t)=(5t2+7t)i+(1-t)j+((4/3)t3+6)k

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