Question

A sonic ranger monitors a cart that moves along a track, monitoring with audible clicks that occur every 28 ms. A computer analyzes the data obtained and gives a best fit function of: x(t) = 0.0012 m/s t + 0.000106 m/s2 t2 - 5e-06 m/s3 t3 (a) At what time does the cart reach its farthest distance from the first sensor? (b) How far from the first sensor does this happen? (c) What is the acceleration at that spot?

Answer 1

a )

given x(t) = 0.0012 m/s t + 0.000106 m/s² t² - 5e-06 m/s³ t³    

x(t) = 0.0012 m/s t + 0.000106 m/s² t² - 5 X 10^-6 m/s³ t³    

x(t) = (1200 t + 106 t² - 5 t³) X 10^-6  

dx(t)/dt = 0

d((1200 t + 106 t² - 5 t³) X 10^-6)/dt = 0

( 1200 + 212 t - 15 t² ) X 10^-6 = 0

15 t² - 212 t - 1200 = 0

t = 18.46 sec , - 4.33 sec

b )

x(t) = (1200 t + 106 t² - 5 t³) X 10^-6  

t = 18.46 sec

x(t) = (1200 X 18.46 + 106 X 18.46² - 5 X 18.46³) X 10^-6  

x(t) = 0.02682 m

c )

acceleration d²x(t)/dt²

d(1200 + 212 t - 15 t² ) X 10^-6 /dt

d²x(t)/dt² = ( 212 - 30t ) X 10^-6

d²x(t)/dt² = - 3.418 m/sec²