Question

Displacement, velocity, speed, acceleration

Answer 1

Displacement after 3s

\(f(3)-f(0)\)

\(f(3)=-(3)^3+6(3)^2-12(3)=-9\)
\(f(0)=-(0)^3+6(0)^2-12(0)=0\)

\(-9-0=-9m\)

Answer 2

Good.

And remember that

Velocity = Displacement / Time

Answer 3

Mhm

\(v(t)=-3t^2+12t-12\)

Right?

Answer 4

That is correct

and
\(a(t) = v'(t) = s''(t)\)

Answer 5

For part C

To find out when the direction changes

s' = v(t) = 0

and solve for t

Answer 6

Speed is \(|v(t)|\) right?

Answer 7

Yes

Answer 8

Average speed

\(\dfrac{\Delta s}{\Delta t}=\dfrac{-9}{3}=-3m\)

Answer 9

Speed at 0

\(|-3(0)^2+12(0)-12|=12\)

Answer 10

Speed at 3

\(||-3(3)^2+12(3)-12|=3\)

Answer 11

\(\Delta d\) not \(\Delta s\) but yes that would be the average velocity for part A haha

Answer 12

I just used the variable for the s function and t time

Answer 13

Show me the letters papa

Answer 14

oh oh, makes sense

I was think of d for displacement

Answer 15

The speed you calculated at t = 0, 3 are correct

Answer 16

Neat

\(a(t)=-6t+12\)

Acceleration at 0:

\(a(0)=-6(0)+12=12m/s^2\)


Acceleration at 3:

\(a(3)=-6(3)+12=-6m/s^2\)

Answer 17

Bravo

Answer 18

ty ty, I would love to thank AZ for the brain support

Answer 19

yw yw, thank you for being the ideal question asker