Question

Displacement, velocity, speed, acceleration part 2

Answer 1

We want v(t)=0

Finding v(t)

\(v(t)=3t^2-18t+24\)

Answer 2

\(3t^2-18t+24=0\)

\(3(t-2)(t-4)=0\)

So we have at \(t=2,4\)

Then to find acceleration, are 2 and 4 on the right track? ;b

Answer 3

so set it equal to 0 and solve

and then we'll evaluate what the acceleration is at those times when velocity = 0

Answer 4

On the right track, champ!

Answer 5

<3


Acceleration:

\(a(t)=6t-18\)


When t is 2 and 4

\(a(2)=6(2)-18=-6\)

\(a(4)=6(4)-18=6\)

Answer 6

\[acceleration =\frac{ dv }{dt }=6t-18\]

Answer 7

at t=2 ,it is retardation

Answer 8

XDD

Answer 9

haha and then similarly for part B, just set a(t) = 0 and then determine the speed at those times

Answer 10

Now for speed when acceleration is 0

\(6t-18=0\)

\(t=3\)

Then to speed

\(|v(t)|=|3(3)^2-18(3)+24|=3m/s\)

Answer 11

splendid

Answer 12

For total distance

\(|s(3)-s(2)|+|s(2)-s(1)|+|s(1)-s(0)|\)?

Or is there another strat

Answer 13

https://www.desmos.com/calculator/9vglnfohey
find distance from 0 to 2 and 2 to 3,then add

Answer 14

since v(t) = 0 at t = 2,4
Those are the time when it changes directions

that means the two intervals we're going to have from t = 0 to t =3 is going to be
0 to 2 and 2 to 3

Answer 15

from the graph we can conclude s=20+2=22

Answer 16

Otherwise, the other strategy once you cover integrals is

\[\text{distance travelled} = \int\limits_{a}^{b}~|v(t)|~dt\]

where a and b are the time intervals

Answer 17

and displacement would be

\[\text{displacement} = \int\limits_{a}^{b}~v(t)~dt\]

Answer 18

ty ty

Answer 19

yw yw

I will add that when you're integrating, you have to separate the bounds whenever it changes direction similar to what we did so you would have two integrals to solve
and it would give us the same answer

\[\text{distance} = \int\limits_{0}^{2}~(3t^2−18t+24)~dt + \int\limits_{2}^{3}~(3t^2−18t+24)~dt\]

\(\text{distance} = 20 + 2 = \boxed{22}\)