Find average acceleration and velocity.

asked by @jakee5 • almost 2 years ago • Physics • 5 pts

Can you please explain the approach to finding average acceleration and velocity (the last two parts of the problem below). enter image description here

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1 answer

AVERAGE VELOCITY

To find the average velocity, we can use the definition:

v_av = Delta x/Delta t

Note that I'm using bold to denote vectors. In this case, Delta x is the DISPLACEMENT, not the distance traveled. Moving from point A to point C, the displacement is 80 m (the diameter of the semicircular part of the track). The displacement is to the left of the figure, which is in the -x direction according to the coordinate system in the figure (it's hard to see the coordinates, but they're there). Noting that the i unit vector is in the +x direction, so the -i unit vector is in the -x direction, the displacement of the runner is

Delta x = -(80 m)i

Now, we're told it takes the runner 15 s to go from point A to point B and another 20 s to go from point B to point C, so

Delta t = 15 + 20 = 35 s

This means that the average velocity is

v_av = -(80 m)i / 35 s = -(2.29 m/s)i

AVERAGE ACCELERATION

The average acceleration is, by definition

a_av = Delta v / Delta t

where Delta v is the change in VELOCITY, not speed. At point A, the velocity is

vA = (8 m/s)j

or 8 m/s in the +y direction (up). At point B, the velocity is

vB = -(7 m/s)i

or 7 m/s in the -x direction (left). At point C, the velocity is

vC = -(6 m/s)j

or 6 m/s in the -y direction (down). So, we need to find both Delta v and Delta t from points A to B (which we'll call "trip 1"), and from points B to C (which we'll call "trip 2"). From A to B:

Delta v1 = -(7 m/s)i - (8 m/s)j

Delta t1 = 15 s

From B to C:

Delta v2 = -(6 m/s)j - [-(7 m/s)i] = (7 m/s)i - (6 m/s)j

Delta t2 = 20 s

So, the average accelerations are:

a1 = [-(7 m/s)i - (8 m/s)j] / 15 s = -(0.47 m/s^2)i - (0.53 m/s^2)j

a2 = [(7 m/s)i - (6 m/s)j] / 20 s = (0.35 m/s^2)i - (0.4 m/s^2)j

answered by @doug • almost 2 years ago