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A particle moves along \(x\)- axis in such a way that its coordinate \((x)\) varies with time \((t)\) according to the expression \(x\) = \(2 - 5t + 6t^{2}\) meters. What is the initial velocity of the particle?

A. \(-5 m/s\)


B. \(-3 m/s\)


C. \(6 m/s\)


D. \(3 m/s\)


Please help me to find the answer.

Note by Sahba Hasan
1 year, 3 months ago

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Initial velocity = first derivative at (t=0) = d/dt(2-5t + 6t^2) at (t=0) = ... Pi Han Goh · 1 year, 3 months ago

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@Pi Han Goh Can you please elaborate, i didn't understand you Sir @Pi Han Goh Sahba Hasan · 1 year, 3 months ago

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@Sahba Hasan what is the derivative of x = 2-5t + 6t^2?

If you substitute t = 0 for the answer you found previously, what does the value you found mean? Pi Han Goh · 1 year, 3 months ago

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Comment deleted Oct 07, 2015

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@Sahba Hasan It's 12t- 5, answer is -5 Pi Han Goh · 1 year, 3 months ago

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Comment deleted Oct 07, 2015

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@Sahba Hasan d/dt(2) = 0 Pi Han Goh · 1 year, 3 months ago

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@Pi Han Goh You got my mistake, i forgot to find derivative of the constant 2. Sahba Hasan · 1 year, 3 months ago

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