Question

A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 s later 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?

Answer 1

The average velocity for three consecutive two-minute intervals is 5.5 m, 7.2 m, and 8.8 m respectively.

How calculate Average velocity?

The average velocity can be calculated by dividing total displacement by the total time.

$\overline{v}={\dfrac{\Delta x}{\Delta t}}$

Where,

$v$ = average velocity

${\Delta x}$    = displacement

${\Delta t}$ = change in time

For First interval,

Displacement  = 25.6 - 14.4 = 11.2 m

Time  = 2 s

So,

$\overline{v}={\dfrac{11.2 }{2}}\\\\\overline{v}= 5.5 \rm \ m/s$

For the second interval,

$\overline{v}={\dfrac{ 14.4}{2}}\\\\\overline{v}= 7.2 \rm \ m/s$

For the third interval,

$\overline{v}={\dfrac{ 17.6}{2}}\\\\\overline{v}= 8.8 \rm \ m/s$

Therefore, the average velocity for three consecutive two-minute intervals is 5.5 m, 7.2 m, and 8.8 m respectively.

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