A particl leaves the orgin with an lintial veloity ` vec u = (3 hat I ) ms &(-1)` and a constant acceleration ` vec a= (-1.0 hat i-0 5 hat j)ms^(-1). Its velocity ` vec v` and positvion vector ` vec r` when it reaches its maximum x-coosrdinate aer .

A particle leaves the origin with an lintial veloity vec u = (3 hat i) and a constant acceleration vec a= (-1.0 hat i-0 5 hat j)ms^(-1) . Its velocity vec v and position vector vec r when it reaches its maximum x-coordinate aer .

A particle leaves the origin with an lintial veloity vec u = (3 hat i) and a constant acceleration vec a= (-1.0 hat i-0 5 hat j)ms^(-1) . Its velocity vec v and position vector vec r when it reaches its maximum x-coordinate what are a. its velosity b. its position vector

vec a = (hat i + hat j + hat k), vec a * vec b = 1 and vec a xxvec b = hat j-hat k. Then vec b is

For a body, angular velocity (vec(omega)) = hat(i) - 2hat(j) + 3hat(k) and radius vector (vec(r )) = hat(i) + hat(j) + vec(k) , then its velocity is :

For a body, angular velocity (vec(omega)) = hat(i) - 2hat(j) + 3hat(k) and radius vector (vec(r )) = hat(i) + hat(j) + vec(k) , then its velocity is :

If vec r*hat i=vec r*hat j=vec r*hat k and |vec r|=3, then find the vector vec r .

If vec a = 5 hat i - hat j - 3 hat k and vec b = hat i + 3 hat j - 5 hat k , then show that the vectors vec a + vec b and vec a - vec b are orthogonal

The position vector of a particle is vec( r ) = ( 3 hat( i ) + 4 hat( j )) metre and its angular velocity vec( omega) =(hat( j)+ 2hat( k )) rad s^(-1) then its linear velocity is ( in ms^(-1) )

The position vector of a particle is vec( r ) = ( 3 hat( i ) + 4 hat( j )) metre and its angular velocity vec( omega) =(hat( j)+ 2hat( k )) rad s^(-1) then its linear velocity is ( in ms^(-1) )