Moving blocks on an incline - relative motion

1. Sep 14, 2010

1. The problem statement, all variables and given/known data
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: aB=52.5m/s2
b) Determine the velocity of B at t = 3.0 seconds. Answer: vB=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.

2. Relevant equations

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

v(t) = v0 + at

3. 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/ss. 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.

Last edited: Sep 14, 2010
2. Sep 14, 2010

collinsmark

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

Last edited: Sep 14, 2010