Question

A person starts to run around a square track with sides of 50 m starting in the bottom right corner and running counter-clockwise. Running at an average speed of 5 m/s, the jogger runs for 53.80 seconds. What is the difference between the magnitude of the person's average velocity and average speed in m/s? (please provide detailed explanation)

Answer 1

v_s = average speed = 5 m/s

t = time of travel = 53.80 s

distance travelled is given as

D = v_s t

D = 5 x 53.80 = 269 m

after travelling distance of 269 m , the person is at point P

hence displacement is given as

AP = sqrt(AB² + BP²)

AP = sqrt(50² + 19²)

AP = 53.5 m

average velocity is given as

V = (AP)/t = 53.5/53.8 = 0.994 m/s

difference in magnitude of average speed and average velocity is given as

difference = average speed - average velocity = 5 - 0.994 = 4.006 m /s