Moving blocks on an incline - relative motion

[JJBladester]

Homework Statement

Block A starts moving downward to the left at a constant acceleration = 80mm/s². At the same time block B moves along block A to the right at an acceleration of 120mm/s² relative to block A.

a) Determine the acceleration of block B. Answer: a_B=52.5m/s²
b) Determine the velocity of B at t = 3.0 seconds. Answer: v_B=157.5m/s²

Note: I think the professor didn't convert between mm's and m's when he obtained the answers. Looks like he wrote the answers in meters by accident.

Homework Equations

a_B/A = acceleration of B with respect to A
a_B = acceleration of B (w.r.t. fixed origin)
a_A = acceleration of A (w.r.t. fixed origin)

v(t) = v₀ + at

The Attempt at a Solution

\left|a_{A}\right|=80mm/s^{2}=.08m/s^{2}
a_{A_{x}}=-a_{A}cos(\theta)\approx-.0274m/s^{2}\hat{i}
a_{A_{y}}=-a_{A}sin(\theta)\approx-.0752m/s^{2}\hat{j}

a_{B/A}=.12m/s^{2}\hat{i}+0\hat{j}

a_{B}=a_{A}+a_{B/A}=(-.0274+.12)\hat{i}+(-.0752+0)\hat{j}
\left|a_{B}\right|=\sqrt{(-.0274+.12)^{2}+(-.0752+0)^{2}}\approx.119m/s^{2}

My answer of .119m/s² is not the correct answer of 52.5m/s^s. Where did I go wrong?

I know the answer to the second part (b) of the question is a simple application of v(t) = v₀ + at.

 

[collinsmark]

Homework Statement

Block A starts moving downward to the left at a constant acceleration = 80mm/s². At the same time block B moves along block A to the right at an acceleration of 120mm/s² relative to block A.

a) Determine the acceleration of block B. Answer: a_B=52.5m/s²
b) Determine the velocity of B at t = 3.0 seconds. Answer: v_B=157.5m/s²

[...snip...]

\left|a_{B}\right|=\sqrt{(-.0274+.12)^{2}+(-.0752+0)^{2}}\approx.119m/s^{2}

My answer of .119m/s² is not the correct answer of 52.5m/s^s. Where did I go wrong?

Given the way the problem statement was written above, I think your answer is correct.