Question

At time t minutes, the position of a body moving along the s-axis is s= t cubed - 27t squared + 200t feet. a. Find the body's acceleration each time the velocity is zero.

Answer 1

Explanation:

Given the expression for the position of a body moving along the s-axis expressed as;

`s = t^3-27t^2+200t`

The acceleration of the body function will be gotten by finding the second derivative of the displacement function as shown;

`(ds)/(dt) = 3t^2-2(27)t+200\\ (ds)/(dt) = 3t^2-54t+200\\\\\\a = (d^2s)/(dt^2) = 2(3)t-54\\a = (d^2s)/(dt^2) = 6t-54\\`

First we need to get the time taken when the velocity is zero.

`Given \ v(t) = (ds)/(dt) = 3t^2-53t+200\\at \ v(t) = 0\\0 = 3t^2-54t+200\\`

factorize;

t = 54±√54²-4(3)(200)/2(3)

t = 54±√2916-2400/2(3)

t = 54±√516/6

t = 54±22.72/6

t = 54+22.72/6 and 54-22.72/6

t = 12.79s and t = 5.21s

Get the acceleration;

at t = 12.79

a = 6t-54

a = 6(12.79)-54

a = 76.74-54

a = 22.74m/s²

at t = 5.21

a = 6t-54

a = 6(5.21)-54

a = 31.26-54

a = -22.74m/s²

Hence the body's acceleration each time the velocity is zero are 22.74m/s² and -22.74m/s²