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in 1992 Maurizio damilano ,of italy walked 29752m in 2.00h( a) calcula damilano's average in m/s. suppose damilano slows down to 3.00m/S at the midpoint in this journey but then picks up the pace and accelerate to the speed calculated in (a).it takes damilano 30.0 s to accelerate .find the magnitude of the average acceleration during this time interval ​

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Answers:

a) [tex]4.132 m/s[/tex]

b) [tex]0.377 m/s^{2} [/tex]

Explanation:

a) Average velocity [tex]V[/tex] is expressed by the following equation:

[tex]V=\frac{d}{t}[/tex]

Where:

[tex]d=29752 m[/tex] is Maurizio's displacement

[tex]t=2h \frac{3600 s}{1 h}=7200 s[/tex] is the time

[tex]V=\frac{29752 m}{7200 s}[/tex]

Hence:

[tex]4.132 m/s[/tex]

b) Average acceleration [tex]a_{ave}[/tex] is the variation of velocity [tex]\Delta V[/tex] over a specified period of time [tex]\Delta t[/tex]:

[tex]a_{ave}=\frac{\Delta V}{\Delta t}}[/tex]

Where:

[tex]\Delta V=V-V_{o}[/tex] being [tex]V_{o}=3 m/s[/tex] the initial velocity and [tex]V=4.132 m/s[/tex] the final velocity

[tex]\Delta t=3 s[/tex]

Then:

[tex]a_{ave}=\frac{4.132 m/s - 3 m/s}{3 s}[/tex]

[tex]a_{ave}=0.377 m/s^{2} [/tex]

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