(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Block A starts moving downward to the left at a constant acceleration = 80mm/s^{2}. At the same time block B moves along block A to the right at an acceleration of 120mm/s^{2}relative to block A.

a) Determine the acceleration of block B.Answer: a_{B}=52.5m/s^{2}

b) Determine the velocity of B at t = 3.0 seconds.Answer: v_{B}=157.5m/s^{2}

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

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) = v_{0}+ at

3. The attempt at a solution

[tex]\left|a_{A}\right|=80mm/s^{2}=.08m/s^{2}[/tex]

[tex]a_{A_{x}}=-a_{A}cos(\theta)\approx-.0274m/s^{2}\hat{i}[/tex]

[tex]a_{A_{y}}=-a_{A}sin(\theta)\approx-.0752m/s^{2}\hat{j}[/tex]

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

[tex]a_{B}=a_{A}+a_{B/A}=(-.0274+.12)\hat{i}+(-.0752+0)\hat{j}[/tex]

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

My answer of .119m/s^{2}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) = v_{0}+ at.

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# Homework Help: Moving blocks on an incline - relative motion

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