The speed of an object moving along the x axis is given by the equation v = a*t - b*t^3.?

Question

(if a = 27 m/s^2 and b = 3.3 m/s^4)

What is the instantaneous acceleration at the highest displacement point?

Answer 1

v = at - bt^3

integrate v with respect to t to get the displacement

x = x0 + (at^2)/2 + (bt^4)/4.......where x0 is the constant of integration

the stationary points are when the velocity is zero

v = 0 = at - bt^3 = t(a - bt^2)

ie when t = 0 or √(a/b) or -√(a/b)

disregard -√(a/b) since its negative

t=0 gives x=x0

and

t=√(a/b) gives x > x0 since (at^2)/2 + (bt^4)/4 > 0 for any t value

so t=√(a/b) gives the highest displacement point

differentiate v with respect to t to get the acceleration

acc = a - 3bt^2

plug in t=√(a/b) gives

acc = a - 3b(a/b) = -2a = -54m/s²

the answer is negative so this means it is decelerating

,.,.,