Please help witht this question?

Question

An object is moving horizontally. The object’s displacement d in meters at t seconds is given by d(t) = -18 + 27t – 10t^2 + t^3. At what time is the acceleration equal to zero? (Answer to two decimal places)

a) 3.33 s
b) 4.79 s
c) 1.88 s
d) 1.00 s

Answer 1

Ok, let's remember some physics :

first derivate of position relative to time = speed(t)

second derivate of position relative to time = acceleration(t)

d(t) = -18 + 27t - 10t^2 + t^3

Let's derivative :

d'(t) = 27 - 20t + 3t^2

d''(t) = -20 + 6t >>> acceleration

the problem says it must be equal to 0 m/s^2

0 = 6t - 20

t = 10 / 3 seconds

t = 3.33 seconds

Hope that helps

Answer 2

2nd differential of distance wrt time gives acceleration.

let distance be s
so
s= -18 + 27t – 10t^2 + t^3
ds/dt = 27 - 20t + 3t^2
d^2s/ dt^2 = -20 + 6t

-20+ 6t = 0
t= 20/6 = 10/3 =3.33 secs

Answer 3

d(t) = -18 + 27t – 10t^2 + t^3
v(t) = d'(t) = 27 -20t +3t^2
a(t) = d"(t) = -20 +6t

a(t) = 0 = -20 +6t; t =20/6 = 3.33

by the way, the units would be meter/second^2

Answer 4

a = d'' = 6 t - 20
a =0 then 6t = 20 then t = 3.33

Answer 5

1.00

-18 + 27*1 - 10*1^2 + 1^3 = -18 + 27 - 10 + 1 = 0

Answer 6

d' = -20t + 3t^2
d'' = -20 + 6t

6t - 20 = 0
6t = 20
t = 20/6