2 dimensional motion problem

Question

Starting from rest at point A, you ride your motorcycle north (the direction) to point B 90.0 m away, increasing speed at a steady rate of 1.55 m/s2. You then gradually turn toward the east (the direction) along a circular path of radius 40.0 m at constant speed from B to point C, until your direction of motion is due east at C. You then continue eastward, slowing at a steady rate of 1 m/s2 until you come to rest at point D.
(a) What is your average velocity and acceleration for the trip from A to D?

(b) What is your displacement during your trip from A to C?

(c) What distance did you travel for the entire trip from A to D?

Answer 1

This problem requires that we determine velocities, coordinates, and times associated with reaching each of the points A, B, C, and D. For accelerated motion we can use the following equations.

(1) d = ½(a)t^2 + v0*t

(2) v = v0 + a*t.

Let’s arbitrarily place point A at the origin of a coordinate system, with the y-axis along the South-North direction and the x-axis along the West-East direction.

1.  Uniform Accerlation

So at time zero, we are at point A with coordinates (0m, 0m). The problem says that we move to point B using uniform acceleration of 1.55 m/s2. Since B is 90 m north of A, it’s coordinates are (0m, 90m). We can calculate the time to reach point B using equation 1. The initial velocity v0 is 0 m/sec.

90m = ½(1.55m/s^2)*t^2. So t^2 = 90m/(0.5*1.55m/s^2) = 116.12 s^2, t = 10.78s. The final velocity at reaching point B is v= 0 + 1.55m/s^2 * 10.78s = 16.70 m/sec.

2.  Uniform Circular Motion

At Point B we begin a turn to the east. Our path is along a circle of radius 40m, and is one quarter of a circle. So at point c we are at coordinates (0+40m, 90+40m) or (40m, 130m). The problem statement is that our speed remains constant during the turn so during the turn we travel at 16.70 m/s up to reaching point C. The path length is one fourth of the circumference of a circle with r=40. So from B to C , we travel d = 0.25 * 2*pi*40m = 62.83m. The time required is t = d/v = 62.83m/(16.70m/s) = 3.76s. So we reach C at time 3.76s + 10.78s = 14.54 seconds.

3.  Uniform Acceleration (deceleration)

Finally we travel from C to D at a constant acceleration. Actually we decelerate to a final velocity of zero. We can determine the time required using equation (2) 0m/sec = 16.7m/sec + (-1m/s^2)*t.

t = 16.7m/s/(-1m/s^2) = 16.70 seconds. So we reach D at time 14.54 + 16.70 or t = 31.24s. Substituting this time into equation 1 tells us that we travel 139.4m east while slowing down. Adding this displacement to our coordinate for C tells us that  D coordiantes = (179.4m, 130m)

Points Time (s)    Velocity (m/s)    Coordinates (m)

A         0               0                  (0, 0)

B         10.78       16.70 North     (0, 90)

C         14.54       16.70 East     (40, 130)

D         31.24          0               (179.4, 130)

Now to answer the questions.

A. Since the velocity at the start and end are both zero, the average acceleration is also zero. Our average speed for traveling from A to D requires finding displacement. Using Pythagorean Theorem. d^2 = 179.4^2 + 130^2. So d = 221.5m. The angle is given by arctan (179.4/130) = 54.1 degrees. So our average velocity is 221.5m/31.24s or 7.09m at 54.1 degrees east of North.

B.  We can find displacement from A to C using the Pythagorean theorem. d^2 = 40^2 + 130^2. So d = 136.0m. Angle is actan (40/130) = 17.10 degrees. Displacement Is 136.0m at 17.10 degrees east of North.

C. To find distance traveled from A to D, add distances for each leg. 90m + 62.8m + 139.4m = 292.2m.