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Turner’s treadmill starts with a velocity of

−2.7 m/s and speeds up at regular intervals

during a half-hour workout.After 33 min, the

treadmill has a velocity of −7.1 m/s.

What is the average acceleration of the

treadmill during this period?

Answer in units of m/s

...?

Verified Answer :

syed514

33
60 

--- 

1980

33*60sec = 1980 sec for comparison

carlosego

By definition, we have to:

\(vf = a * t + vo
\)

Where,

vf: final speed

vo: initial speed

t: time

a: acceleration

Clearing the acceleration we have:


\(a = \frac{vf-vo}{t}
\)

When replacing values we have to take into account two things:

1) Assume that the given speeds are positive

2) The time must be in seconds

The time in seconds is:

\(t = 33 * 60

t = 1980
\)

Then, replacing values we have:


\(a =\frac{7.1-2.7}{1980}
\)

\(a = 0.002 \frac{m}{s ^ 2}
\)

Answer:

The average acceleration of the treadmill during this period is:

\(a = 0.002 \frac{m}{s ^ 2}
\)

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