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Physics 7 Online
OpenStudy (anonymous):

A car moving at a speed of 20 m/s, then accelerates uniformly at 1.8m/s2 until it reaches a speed of 25m/s. What distance does it travel during the time it is accelerating?

OpenStudy (anonymous):

@douglaswinslowcooper

OpenStudy (anonymous):

@NERD85176

OpenStudy (anonymous):

\[a=\frac{ V_{f}-V_{i} }{ t }\] So we have a, Vi and Vf. So we need t: \[t=\frac{ V_{f}-V_{i} }{ a }\] \[t=\frac{ 25\frac{ m }{ s }-20\frac{m }{ s } }{ 1.8\frac{ m }{ s^{2} } }\]

OpenStudy (anonymous):

t=\[\frac{ 25 }{ 9 }\]\[x_{f}-x_{i}=v_{0}t+\frac{ 1 }{ 2 }at ^{2}\] \[d=20(\frac{ 25 }{ 9 })+\frac{ 1 }{ 2 }(1.8)(\frac{ 25 }{ 9 })^{2}\]

OpenStudy (anonymous):

d=64.34461806 m I got those formulas from the internet

OpenStudy (anonymous):

Ohh I think I had a mistake it is d=125/2=62.5. I

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