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A bicyclist brakes while approaching a stop sign, accelerating uniformly as the bike slows to a stop. The bicyclist presses the brakes when 3.3m away from the stop sign and it takes the bike 2.4s to stop at the stop sign. What was the acceleration of the braking bicyclist?

 Nov 14, 2021
 #1
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Can't say without knowing initail velocity .

 

Example     travelling  10 m/s

 x =     xo + vo t - 1/2 a t^2

            3.3 + 10(2.4) - 1/2 a(2.4^2 )      a (decceleration) = 9.48 m/s^2

 

Example 2    travelling  20 m/s

               3.3  + 20 (2.4) - 1/2 a(2.4)^2    a = 17.81 m/s^2

 

 

Soooo  you need to know the initial velocity to know how fast they slowed down to a stop  ( the deccelaration)

 Nov 14, 2021
 #2
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Solution:

 

You can derive the initial velocity from the average velocity via the traveled distance where the breaking begins to the stop sign:

 

\(\text {The average velocity is } \dfrac{3.3 \tiny \;meter \hspace{-.1em}s} {2.4 \;s} = 1.375 \tiny \dfrac {m}{s}, \normalsize \text { then multiply by (2) for the initial velocity } (2 *1.375) = 2.75\tiny \dfrac {m}{s} \\ \text {Then }\\ a = \dfrac{v - u}{t} \hspace{1em}| \hspace{1em} \small \text {a = acceleration, v = final velocity, u = initial velocity, t = time} \\ \dfrac{0 - 2.75 \tiny m/s}{2.4s} = -1.5625 \; ms^{-2} \Leftarrow \text {negative acceleration} \)

 

GA

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Note: “Help” is spelled with an “e” not an “a” so, reserve the baby-talk for babies. Most babies do not know how to read....

 Nov 15, 2021

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