From t = 0 to t = 5.00 min, a man stands still, and from t = 5.00 min to t = 10.0 min, he walks briskly in a straight line at a constant speed of 2.3 m/s.

(a) What is his average velocity vavg in the interval 2.000 min to 7.90 min?
I found this to be 1.15 m/s

(b) What is his average acceleration aavg in the interval 2.00 min to 7.90 min?
How do I go about solving this?
Would I do 1.15m/s divided by 354 seconds?

actually should I do 2.3 m/s divided by 343 seconds to get .00599 m/s^2?

To find the average acceleration, you need to calculate the change in velocity and divide it by the time interval. In this case, the man is standing still from t = 0 to t = 5.00 min, so his velocity during that interval is 0 m/s.

To find his velocity from t = 5.00 min to t = 10.0 min, you know that he walks briskly at a constant speed of 2.3 m/s. So his velocity during that interval is also 2.3 m/s.

Now, to find the average acceleration during the interval 2.00 min to 7.90 min, you can use the formula:

aavg = (v2 - v1) / (t2 - t1)

where v2 and v1 are the final and initial velocities, and t2 and t1 are the final and initial times, respectively.

v2 = 2.3 m/s (final velocity from t = 5.00 min to t = 10.0 min)
v1 = 0 m/s (initial velocity from t = 0 to t = 5.00 min)
t2 = 7.90 min (final time for the interval 2.00 min to 7.90 min)
t1 = 2.00 min (initial time for the interval 2.00 min to 7.90 min)

To convert minutes to seconds, you need to multiply by 60.

t2 = 7.90 min * 60 s/min = 474 s
t1 = 2.00 min * 60 s/min = 120 s

Substituting these values into the formula, you get:

aavg = (2.3 m/s - 0 m/s) / (474 s - 120 s)
= 2.3 m/s / 354 s
≈ 0.0065 m/s^2

So the average acceleration in the interval 2.00 min to 7.90 min is approximately 0.0065 m/s^2.