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# Calculus help? This is due in an hour?? What did I do wrong?

I got v(t)=-t^2+10t-24

a(t)=-2t+10

Moving right: (0,6),(4,7)

Moving left: (6,4)

Speeding up: (0,6),(5,4)

Slowing down: (6,5),(4,7)

I set both equation equal to 0 to get the 2 points but my teacher told me the solutions to them were wrong?

BY SOLUTIONS DOES HE MEAN PROCESS BY WHICH I SOLVED IT WAS WRONG? I used the quadratic equation??

### 5 Answers

- AlvinLv 41 month agoFavorite Answer
Due to the definition of speed, but both of your answers have gotten

the 2nd part wrong due to some tricky issue.

velocity and speed are different.

For part ii, your answers indicate when the velocity is

increasing, but not the speed.

Speed = Magnitude of velocity

so both are your answer so far are incorrect

technically for part ii.

Definition of Speeding Up

Note: Velocity is a signed variable

when going to the right,

+ increasing positive velocity meaning increasing speed to

the right

when going to the left,

decreasing negative velocity meaning increasing speed to the left

So Speed = | Velocity |

so when you are going left .

Speed = - (-t^2 +10t -24 ) = t^2 -10t -24

Rate of speed = 2t -10

so between 0 and 4

Rate of speed = (2 (0 to 4) ) -10 = negative

so slowing down

the rate of the speed is negative

is negative so you are slowing down but going to the left

From 6 to 7 ,

rate of speed = 2t-10 = 2 *(6 to 7) -10 = positive

so you are speeding up.

From 4 to 6

speed (t) = -t^2 +10t -24

rate of speed = -2t + 10

so from 4 to 5

rate of speed is positive

rate of speed = -2(4 to 5) + 10 = positive

from 5 to 6

rate of speed =-2(5 to 6) +10 = negative

speed (t) =

so the correct answer to part ii

Increasing speed

(4,5) (6,7)

Decreasing speed

(0,4) (5,6)

- PinkgreenLv 71 month ago
The direction of the increasing values on the x-axis is

considered "right".

s(t)=-(t^3)/3+5t^2-24t+8

=>

s(0)=8

s(4)=-29.3333...

s(5)=-28.6666...

s(6)=-28

s(7)=-29.3333...

s'(t)=-t^2+10t-24=-(t-4)(t-6)---[speed]

s'(t)=0=>t=4 or t=6

s"(t)=-2t+10------[acceleration]

s"(4)=-2(4)+10=2>0=>s'(4,-29.333..) is a min.

s"(6)=-2(6)+10=-2<0=>s'(6,-28) is a max.

s"(5)=0=>s'(5,-28.666..) is a pt. of inflection.

(i)

Thus, (0,4) & (6,7) in which the particle moves to the left;

(4,5) & (5,6) moves to the right.

(ii)

In (0,4) & (4,5);

s'(0)=-24, s'(4)=0, s'(5)=1. The speed is increasing in math-point of view.

s"(0)=10,s"(4)=2,s"(5)=0. The particle is decelerated to the left, but accelerated to the right.

In (5,6) & (6,7);

s'(5)=1,s'(6)=0,s'(7)=-3. The speed is decreasing in math-point of view

s"(5)=0,s"(6)=-2, s"(7)=-4. The particle is accelerated to the right, but decelerated to the left.

s"(t)=-2t+10=>S"(t) is a linear function of t with the slope=-2=>s"(t) always decreases

in 0=<t<=7.

- Wayne DeguManLv 71 month ago
v(t) = -t² + 10t - 24...as you correctly state.

Equating to zero we get:

-t² + 10t - 24 = 0

or, -(t² - 10t + 24) = 0

so, -(t - 6)(t - 4) = 0

Hence, the particle is stationary when t = 4 and again when t = 6

Therefore, we need to examine what happens for 3 intervals

i.e. when 0 < t < 4, 4 < t < 6 and t > 6

Graphically, the function v(t) is a ∩ - shaped parabola

Hence, for 0 < t < 4 the function is below the horizontal axis, i.e. negative => moving left for (0, 4)

For 4 < t < 6 the function is above the axis, i.e. positive => moving right for (4, 6)

Finally, for t > 6 the function is below the horizontal axis, i.e. negative => moving left for (6, 7)

a(t) = -2t + 10...as you correctly state

Again, equating to zero we get:

-2t + 10 = 0

so, t = 5...is when the particle's velocity is constant, i.e. zero acceleration.

Graphically, this is a line with negative gradient

Hence, for 0 < t < 5...a(t) is positive => speeding up for (0, 5)

Then, for 5 < t < 7...a(t) is negative => slowing down for (5, 7)

Note: we could have used other techniques, but considering the graphical aspect gets you to think about the situation as a picture and see what the behaviour is.

:)>

- ted sLv 71 month ago
sorry , all your answers are wrong....key values for v are x = 0 , 4, 6, 7

v < 0 on (0,4) U ( 6 , 7) and a > 0 on (0,5)

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- AlanLv 71 month ago
You don't understand interval notation

The interval is the range

it is not the endpoints

(0,6) to (4,7) is time interval (0,4)

when v(t) is positive you are going right

when v(t) is negative you are going left .

when is -t^2 +10t -24 crossing 0

factor it to determine this

-1(t^2 -10t+24) = -(t-6)(t-4)

so you have 3 number

3 negative make it negative

when t < 4 , -1*(negative)(negative) = negative

when 4< t< 6, -1*(negative)(positive) = positive

when t> 6 , -1(positive)(positive) = -negative

Moving right v(t) =-t^2 +10t -24 is positive

(4,6)

moving left

when y = -t^2 +10t - 24 is negative

graph you will see that

(0,4) (6,7)

Next question

a(t) = -2t + 10

0= -2t+10

2t =10

t = 5

so when t t<5 ,

a(t) =- negative number greater than -10 + 10 = positive

t> 5

a(t) =negative number less than -10 +10= negative

so speeding up

(0,5)

slowing down

(5,7)