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Math Help - Transpose equation problem

  1. #1
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    Transpose equation problem

    Hi all, I'm having trouble transposing this equation so I can work out when acceleration first reaches zero on a body of a mass
    dv/dt= 6e^-t/2 (2-t) so when

    0= 6e^-t/2(2-t) what does t equal

    i know I need to get rid of the brackets first but that's what's confusing me its been a while
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  2. #2
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    Re: Transpose equation problem

    Quote Originally Posted by jonoliv8 View Post
    Hi all, I'm having trouble transposing this equation so I can work out when acceleration first reaches zero on a body of a mass
    dv/dt= 6e^-t/2 (2-t) so when

    0= 6e^-t/2(2-t) what does t equal

    i know I need to get rid of the brackets first but that's what's confusing me its been a while
    You need to use the zero product principle!

    0=6e^{-\frac{t}{2}}(2-t)

    This gives two different equations

    6e^{-\frac{t}{2}}=0 \quad \text{or} \quad 2-t=0

    Since the exponential is never zero, the first equation has not solutions, but the 2nd equation does ...
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  3. #3
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    Re: Transpose equation problem

    I'm not familiar with that rule, I can see just looking at the equation that when the brackets equal zero ie t=2 then dv/dt=0 but the question asks to transpose it?
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  4. #4
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    Re: Transpose equation problem

    Hello, jonoliv8!

    You're overlooking a simple concept . . .


    a \:=\:\tfrac{dv}{dt} \:=\:6e^{\text{-}\frac{1}{2}t}(2-t)

    \text{Determine when }a = 0.

    We have: . 6e^{-\frac{1}{2}t}(2-t) \;=\;0

    Divide by 6e^{\text{-}\frac{1}{2}t}\!:\;\;2-t \:=\:0 \quad\Rightarrow\quad \boxed{t \,=\,2}


    \text{Note that: }\:6e^{-\frac{1}{2}t} \:=\:\frac{6}{e^{\frac{1}{2}t}} \;\ne\;0
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  5. #5
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    Re: Transpose equation problem

    Damn it, ive been doing assignment questions all day and I've been staring at that equation for2 hours lol, it just takes someone to slap you round the face with the right answer to wake up ha, thanks for the help I can't belive I didn't see that.
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  6. #6
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    Re: Transpose equation problem

    Quote Originally Posted by jonoliv8 View Post
    I'm not familiar with that rule, I can see just looking at the equation that when the brackets equal zero ie t=2 then dv/dt=0 but the question asks to transpose it?
    My guess it you do know the zero product principle.

    For example how would you solve?

    t^2-5t+6=0 \iff (t-2)(t-3)=0 \implies t-2 =0 \quad t-3=0
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