(e) If the wheel rolls along the ground without slipping, the instantaneous velocity of the atoms of the object that are momentarily in contact with the ground is zero. This zero-velocity condition implies that , where is the angular speed of the object, since the instantaneous speed of the contact point is . During the time between 10 s and 15 s, how far did the center of the wheel move, in meters?

distance moved by the center of the wheel during time  to    is  145.32 m

Explanation:

Given that :

The radius of the wheel  R =  28 cm

Initial angular speed of the wheel

=

=

= 100.5 rad/s

At time t = 9 s ;  the angle  rotated by wheel

= 32 π × 9

= 905 rad

At time ; the angular speed is definitely the same and the initial velocity is

However ; after time   ; the angular acceleration of the wheel ∝ = 1.3 rad/s²

At time ; angular speed of wheel

Now; the angle rotated by the wheel during time    to    is expressed as:

From the question; we are being told that :

This zero-velocity condition implies that vCM=ω⁢R, where ω is the angular speed of the object, since the instantaneous speed of the contact point is vCM-ω⁢R.

i.e

then we can say :

acceleration    and

linear distance    indicating   how far did the center of the wheel move

So; distance moved by the center of the wheel during time  to    is ;

= 0.28 × 519

s = 145.32 m

Therefore; distance moved by the center of the wheel during time  to    is  145.32 m

see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west

i believe the answer is a hope it and is correct