The above answer is excellent. But there is a simpler way to see that the angle between speed and acceleration is 45 degrees.
Consider the motion composed out of a uniform translation with uniform horizontal speed `v` and a uniform rotation with uniform angular speed `omega` .
Point P will have the speed `v(= omega*R)` upwards from the uniform rotation, and speed `v` horizontal from the uniform translation. Thus the total speed of point P is `v*sqrt(2) (=omega*R*sqrt(2))` and is directed at 45 degree with the horizontal.
Point P will have the acceleration `a_(cp) = omega^2*R` directed horizontal (towards the center of rotation C) from rotation motion, and `a=0 m/s^2` from linear translation motion. Thus the total acceleration of point P is horizontal.
Therefore the angle between total speed and total acceleration is 45 degree.
Consider the image attached.
The blue arrows represent the velocity of point P. The arrow pointing straight up is the velocity due to rotation and it is equal wR, where w is the angular velocity.
The arrow pointing horizontally is the velocity due to rolling forward. Since the disk will move forward the distance equal to its circumference at the same time as the rotation motion will go through one circle, this velocity will also equal wR.
This means the resultant net velocity, which is the sum of vertical and horizontal components equal in length, will make 45 degrees with the horizontal. It's magnitude will be `sqrt2 wR` .
The green arrows represent the acceleration. The normal (radial) acceleration is the green arrow pointing down (along line PC) and is perpendicular to the net velocity, which is tangential to the trajectory of point P. Point C, the instantaneous contact between the disk and the surface, is the instanteneous center of the curvature of trajectory of point C. Note that `PC = sqrt2 R` .
Therefore, normal acceleration is
`a_n = V^2/R = (2w^2R^2)/(sqrt2 R) = sqrt2 w^2 R`
The tangential acceleration is colinear with the net velocity. If r is the distance between moving point P and point C, then `r=sqrt(2)R`
`a_t = (dV)/dt = w(dr)/(dt) = wr(d(theta))/(dt) = w^2r=sqrt2 w^2R`
`` So the tangential and normal accerleration have the same magnitude, which means the net acceleration will be directed horizontally (making 45 degree angle with either), and also making 45 degree angle with net velocity.
The answer is B, 45 degrees.
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