#Impossible mechanics Question

22 messages · Page 1 of 1 (latest)

narrow cradle
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I don't understand part B?

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woven trellisBOT
narrow cradle
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<@&286206848099549185>

marsh wigeon
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You don't understand the task or you can't find the solution?

narrow cradle
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Don't understand how to do part b at all

narrow cradle
marsh wigeon
# narrow cradle I need full breakdown of the answer

You need to establish what is given and what you are looking for. Start by defining T:
$\newline T+t_{M}=25s+t_{N}\newline$ where $t_{N,M}$ is the time it takes for each train to come to a halt after starting to break.

jagged tigerBOT
marsh wigeon
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The first objective is to calculate the time it takes the second train (N) to stop. This can be done as follows:

$\newline v\left(t\right)=v_{0}+at \newline$
$\newline 0=30ms^{-1}-a_{N}t_{N} \newline$

Solving for $a_{N}$ yields $\newline\newline a_{N}=\frac{30ms^{-1}}{t_{N}}$

$\newline s\left(t\right)=s_{0} + v_{0}\cdot t+\frac{1}{2}at^{2} \newline$
$\newline 975m=30ms^{-1}\cdot25s+30ms^{-1}\cdot t_{N}-\frac{1}{2}a_{N}t_{N}^{2} \newline$

Substituting for $a_{N}$ into $s(t)$:

$\newline 975m=30ms^{-1}\cdot25s+30ms^{-1}\cdot t_{N}-\frac{1}{2}\frac{30ms^{-1}}{t_{N}}t_{N}^{2} \newline$

and after simplifying and solving for $t_{N}$:

$\newline t_{N}=15s \newline$

Rinse and repeat for $t_{M}$ (you can skip actually calculating $t_{M}$ by substituting T).

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This assumes that you are familiar with the equations for linear motion.

jagged tigerBOT
floral blade
woven trellisBOT
marsh wigeon
floral blade
marsh wigeon
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It's not like I spoiled the solution itself.

floral blade
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@narrow cradle you still with us?

narrow cradle
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Nah I need someone to write the answer out very confusing

marsh wigeon
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Oh, well. @narrow cradle Is there anything specific you don't understand?

narrow cradle
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Nah don't understanding how to do the question