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? 

Update:

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

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5 Answers

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  • Alvin
    Lv 4
    1 month ago
    Favorite 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)    

       

     

  • 1 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.

  • 1 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 s
    Lv 7
    1 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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  • Alan
    Lv 7
    1 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) 

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