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.

## Comments

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TopNewestInitial 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 – Sahba Hasan · 1 year, 3 months ago

Can you please elaborate, i didn't understand you SirLog in to reply

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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– Pi Han Goh · 1 year, 3 months ago

It's 12t- 5, answer is -5Log in to reply

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– Pi Han Goh · 1 year, 3 months ago

d/dt(2) = 0Log in to reply

– Sahba Hasan · 1 year, 3 months ago

You got my mistake, i forgot to find derivative of the constant 2.Log in to reply