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)`

Step-by-step explanation:

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

`V=(d)/(t)`

Where:

`d=29752 m` is Maurizio's displacement

`t=2h (3600 s)/(1 h)=7200 s` is the time

`V=(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)=(\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)=(4.132 m/s - 3 m/s)/(3 s)`

`a_(ave)=0.377 m/s^(2)`