Question

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 ​

Answer 1

Answers:

a) $4.132 m/s$

b) $0.377 m/s^{2}$

Explanation:

a) Average velocity $V$ is expressed by the following equation:

$V=\frac{d}{t}$

Where:

$d=29752 m$ is Maurizio's displacement

$t=2h \frac{3600 s}{1 h}=7200 s$ is the time

$V=\frac{29752 m}{7200 s}$

Hence:

$4.132 m/s$

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

$a_{ave}=\frac{\Delta V}{\Delta t}}$

Where:

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

$\Delta t=3 s$

Then:

$a_{ave}=\frac{4.132 m/s - 3 m/s}{3 s}$

$a_{ave}=0.377 m/s^{2}$