CAN SOMEONE HELP ME ON THIS!

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

Question

CAN SOMEONE HELP ME ON THIS!

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

anonymous · Answer

$$\frac{V _{f}-V_{i}}{t}$$There's your formula.  Convert your minutes to seconds, plug in your values, and you're done.

amistre64 · Answer

is should be the slope between the points: (0,-2.7) and (33,-7.1) if i read it right

amistre64 · Answer

which comes to what brandon posted :)

anonymous · Answer

whats the answer so that i know if i have the answer right.

amistre64 · Answer

what answer did you get?

amistre64 · Answer

$$30\ min*\frac{60\ sec}{1\ min}$$
$$30\ \cancel{min}*\frac{60\ sec}{1\ \cancel{min}}$$
$$30*60\ sec=1800\ sec$$

amistre64 · Answer

correction;

33
60
---
1980

33*60sec = 1980 sec for comparison

anonymous · Answer

I got 0.0022222?

amistre64 · Answer

i get:
  (1980,-7.2)
-(  0    ,-2.7)
-------------
  1980, -4.5

slope = -4.5/1980 = -21/(5)1980 = -7/3300

or,  -.002121....

amistre64 · Answer

or something thereabouts ....

amistre64 · Answer

are those spose to be negative signs or just a way to indicate with the number is?