- #1

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I don't understand how to calculate.

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- MHB
- Thread starter Shah 72
- Start date

- #1

- 274

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I don't understand how to calculate.

- #2

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What have you been able to do so far?

-Dan

-Dan

- #3

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V= integration (6t-c) with limits 0 and -4. I got the ans c= -12. But it's wrong and the textbook ans is 6What have you been able to do so far?

-Dan

- #4

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I got c=12 not -12What have you been able to do so far?

-Dan

- #5

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\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)I got c=12 not -12

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

Try to finish it.

-Dan

- #6

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You mean I do s=t^3-ct^2/2\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

Try to finish it.

-Dan

- #7

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You take both the equations = 0 as stationary .\(\displaystyle a = \dfrac{dv}{dt} = 6t - c\)

so \(\displaystyle v = \int_0^t (6t - c) ~ dt = 3t^2 - ct\)

\(\displaystyle v = \dfrac{dx}{dt} = 3t^2 - ct\)

Try to finish it.

-Dan

Then solve the quadratic equation by using b^2-4ac=0. I still don't get the ans

- #8

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Your formula for s (I've been calling it x) is correct. Now you know when the motion of the particle stops, v(t) = 0 and x(t) = -4. SoYou take both the equations = 0 as stationary .

Then solve the quadratic equation by using b^2-4ac=0. I still don't get the ans

\(\displaystyle 0 = 3t^2 - ct\)

and

\(\displaystyle -4 = t^3 - \dfrac{1}{2} c t^2\)

My advice is to solve the top equation for ct, multiply it by t (to get \(\displaystyle ct^2\)) and then plug that into the bottom equation.

-Dan

- #9

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Thank you so much!Your formula for s (I've been calling it x) is correct. Now you know when the motion of the particle stops, v(t) = 0 and x(t) = -4. So

\(\displaystyle 0 = 3t^2 - ct\)

and

\(\displaystyle -4 = t^3 - \dfrac{1}{2} c t^2\)

My advice is to solve the top equation for ct, multiply it by t (to get \(\displaystyle ct^2\)) and then plug that into the bottom equation.

-Dan

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