Step-by-step explanation:
Given the expression for the position of a body moving along the s-axis expressed as;
[tex]s = t^3-27t^2+200t[/tex]
The acceleration of the body function will be gotten by finding the second derivative of the displacement function as shown;
[tex]\frac{ds}{dt} = 3t^2-2(27)t+200\\ \frac{ds}{dt} = 3t^2-54t+200\\\\\\a = \frac{d^2s}{dt^2} = 2(3)t-54\\a = \frac{d^2s}{dt^2} = 6t-54\\[/tex]
First we need to get the time taken when the velocity is zero.
[tex]Given \ v(t) = \frac{ds}{dt} = 3t^2-53t+200\\at \ v(t) = 0\\0 = 3t^2-54t+200\\[/tex]
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
