Recent content by ladkee

I got the right answers, 1.39 & - 80.06, thanks for all ur help, highly appreciated

Okay, for magnitude I did: |a| = sqrt ( (0.24^2) + (-1.37^2) = 1.39 m/s2 and direction I did inverse tangent of (y/x) which is ( -1.37 / 0.24 ) = - 80.06 deg ...which I'm not sure is right

so should it be something like ... a_A = (4*i + (- 3.76*j) ) + ( 0*i + (-1.37*j) ) = 0.24*i - 1.37*j

I'm not sure what you mean by that? ...so are you saying for the x & y components I need to use are (-3.76, -1.37). How do you find the magnitude & direction for acceleration for this particular problem? please help, I need a clear response.

what is the next step to find the magnitude of acceleration for a_A, are you saying use (-3.76, -1.37) as the x&y components?

Homework Statement Block B has acceleration of 4 m/s2... Relative acceleration of block A w/ respect to B is 4 m/s2. Find magnitude & direction of accel for A? Homework Equations a_A = a_B + a_A/B x_A = x_B + x_A/B y_A = y_B + y_A/B The Attempt at a Solution x & y components: -4cos(20) =...

Homework Statement : Block B has acceleration of 4 m/s2... Relative acceleration of block A w/ respect to B is 4 m/s2. Find magnitude & direction of accel for A?[/B] Homework Equations a_A = a_B + a_A/B[/B]The Attempt at a Solution a_A = 4 +4 = 8 m/s2 <-- 8*cos(20) = 7.52 (x-component)...