At time t=0, a particle is located at point (5, 9, 2). It travels in straight line to point (6,1,3), has speed 2 at (5,9,2) and constant acceleration i-8j+k. Find the equation for the position vector r(t) of particle at time t.
-
Since it travels with constant acceleration, you can use the second uniform motion formula
s=ut +1/2at^2
Since this is a vector question, all variables I'm using from now on will be vectors
Let p be the position vector 5i+9j+2k of the initial point.
Since s is displacement, it is given by s=r(t) -p
u is the initial velocity. Direction is from (5,9,2) to (6,1,3)
A vector along u is (6-5)i+(1-9)j+(3-2)k = i-8j+k; whose magnitude is sqrt(1^2+8^2+1^2)=sqrt(66)
Since u is of magnitude 2, u is (i-8j+k) *2/sqrt(66)
Substituting in formula,
r(t) -p=ut +1/2at^2
r(t)=p+ut+1/2at^2
=(5i+9j+2k) +(i-8j+k) *2/sqrt(66) t +1/2 (i-8j+k) t^2
All that's left is to add up the individual components; but by my calculations, that would look complex
Note: this is more of a physics question than a maths one
s=ut +1/2at^2
Since this is a vector question, all variables I'm using from now on will be vectors
Let p be the position vector 5i+9j+2k of the initial point.
Since s is displacement, it is given by s=r(t) -p
u is the initial velocity. Direction is from (5,9,2) to (6,1,3)
A vector along u is (6-5)i+(1-9)j+(3-2)k = i-8j+k; whose magnitude is sqrt(1^2+8^2+1^2)=sqrt(66)
Since u is of magnitude 2, u is (i-8j+k) *2/sqrt(66)
Substituting in formula,
r(t) -p=ut +1/2at^2
r(t)=p+ut+1/2at^2
=(5i+9j+2k) +(i-8j+k) *2/sqrt(66) t +1/2 (i-8j+k) t^2
All that's left is to add up the individual components; but by my calculations, that would look complex
Note: this is more of a physics question than a maths one