(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?

A sonic ranger monitors a cart that moves along a track, monitoring with audible clicks that occur every 6 ms. A computer analyzes the data obtained and gives a best fit function of:

x(t) = 0.00148 m/s t + 0.000136 m/s2 t2 - 5.3e-06 m/s3 t3

1 Answer
A08
Apr 2, 2018

This is what I get.

Explanation:

From the given expression for #x(t)# we note that the movement of the cart is having component of variable acceleration. Also that after some time the direction of motion changes from #+ve# to #-ve# due to #-t^3# term.

(a) As such cart reaches its farthest distance from the first sensor when velocity becomes zero.

#x (t)= 0.00148 t + 0.000136 t^2 - 5.3xx10^(-6) t^3#
#v=dotx = 0.00148 + 0.000136xx2 t - 5.3xx10^(-6)xx3 t^2#
#=>v=dotx = 0.00148 + 0.000272 t - 1.59xx10^(-5) t^2#

Under the given condition

#0 = 0.00148 + 0.000272 t - 1.59xx10^(-5) t^2#

Solving the quadratic using the inbuilt graphics utility and ignoring the #-ve# root as time can not be negative we get

enter image source here

#t=21.447001\ s#

(b) Distance from the first sensor

#s=0.00148 xx21.447001 + 0.000136 (21.447001)^2 - 5.3xx10^(-6) (21.447001)^3#
#s=0.042\ m#

(c) Acceleration #a=dotv = 0.000272 - 3.18xx10^(-5) t#
At the desired location

#a = 0.000272 - 3.18xx10^(-5) xx21.447001#
#a=-0.00041\ ms^-2#

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