Question

X=t^3-3t^2+6t-5 find initial velocity & acceleration

Answer 1

Given :

✏ Equation of position of particle in terms of time is provided...

$\boxed{\sf{\red{x=t^3-3t^2+6t-5}}}$

To Find :

• Initial velocity of particle
• Acceleration of particle

Concept :

$\circ\sf\:velocity(v)=\dfrac{dx}{dt}\\ \\ \circ\sf\:acceleration(a)=\dfrac{dv}{dt}$

Calculation :

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• Initial velocity

$\implies\sf\:v=\dfrac{d(t^3-3t^2+6t-5)}{dt}\\ \\ \implies\sf\:v=3t^2-6t+6\\ \\ \red{\dag\:{\sf{at\:t = 0 \: time,}}}\\ \\ \implies\sf\:v=3(0)^2-6(0)+6\\ \\ \implies\: \boxed{\sf{\purple{v=6\:mps}}}$

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• Acceleration

$\mapsto\sf\:a=\dfrac{d(3t^2-6t+6)}{dt}\\ \\ \mapsto\sf\:a=6t-6\\ \\ \green{\dag\:{\sf{at\:t=0\:time}}}\\ \\ \mapsto\sf\:a=6(0)-6\\ \\ \mapsto\:\boxed{\sf{\orange{a=-6\:ms^{-2}}}}$

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Additional information :

• Velocity and acceleration both are vector quantities.
• Vector quantity has both magnitude as well as direction.