The figure gives the path of a squirrel moving about on level ground, from point
ID: 1550200 • Letter: T
Question
The figure gives the path of a squirrel moving about on level ground, from point A (at time t = 0), to points B (at t = 5 min), C (at t = 10 min), and finally D (at t = 15 min). Consider the average velocities of the squirrel from point A to each of the other three points. Of them, what are the (a) magnitude and (b) angle of the one with the least magnitude and (c) the magnitude and (d) angle of the one with the greatest magnitude? Give the angles in the range (-180°, 180°]
y (m) 50 25 25 -50 A SC (m)Explanation / Answer
It is a 2D motion of the squirrel. Average velocity is given by:
v= Total displacement / Total time
The displacement between two points is the shortest distance between those points.
For point B
Displacement along x axis from point A = x = 15m
Displacement along y axis from point A = y = -30 m
Total displacement = d = (152+302)1/2 = 33.54
Total time = t =5 min
Hence average velocity from A = v = 33.54/5 = 6.71 m/min
angle = w = tan-1(y/x) = tan-1(-30/15) = -63.4 deg
For point C
Displacement along x axis from point A = x = 5m
Displacement along y axis from point A = y = 0 m
Total displacement = d = 5
Total time = t =10 min
Hence average velocity from A = v = 5/10 = 0.5 m/min
angle = w = tan-1(y/x) = tan-1(0/5) = 0 deg
For point D
Displacement along x axis from point A = x = 30m
Displacement along y axis from point A = y = 60 m
Total displacement = d = (602+302)1/2 = 67.08
Total time = t =15 min
Hence average velocity from A = v = 67.08/15 = 4.47 m/min
angle = w = tan-1(y/x) = tan-1(60/30) = 63.4 deg
Hence point B has the highest magnitude of velocity, 6.71m/min (angle =-63.4)
Point C has the smallest magnitude, 0.5 m/min (angle =0)