The position of a cucumber is given by x=at^2-bt^2 where a and b are constants.

Question

The position of a cucumber is given by

x=at^2-bt^2

where a and b are constants. Positive x is to the right and negative x is to the left.

a) At what time t will the cucumber be the farthest to the right? (Your answer will be in terms of a and b)

b) How far away from its initial position will the cucumber be at the time find in part (a)? That is, what is the maximum distance the cucumber travels to the right?

c) What is the cucumber's average velocity over this time interval (i.e. from t=0 to the time find in (a))?

d) Confirm that your answers to parts (a), (b), and (c) have the correct units. (First you'll need to find the units of a and b.)

Explanation / Answer

check your question

i think x should be

x = at-bt^2

( where a,b >0)

units of

x = meter

t = seconds

By equating both sides units

a = meter/seccond

b = meter/second^2

a)

dx/dt = a - 2*b*t

=> dx/dt = 0 will give inflection points

=> a-2*b*t = 0 => t = a/2b seconds

So at time t = a/2b x will be farthest to right

b) x at t= a/2b

=> x = a*a/2b - b*(a/2b)^2 = a^2/2b - a^2/4b

=> x = a^2/4b meters

c) change in x = a^2/4b - 0 = a^2/4b

so avg. velocity = (a^2/4b)/(a/2b) = a/2 meter/second

d) dimensions of a/2b = (meter/second)/(meter/second^2) = second = [T]

dimension of a^2/4b = (meter/second)^2/(meter/second^2) = meter = [L]

dimension of a/2 = meter/second = [L*T^-2]

Which is appropriate.

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