Question

The velocity as a function of time of a moving particle is given by v = α+ βt2 , where α and β are constants and t is time in s. What is the acceleration of the particle at 3 s?​

Answer 1

The acceleration of the particle at 3s is [tex]a = 6 \beta [/tex]

$v = \alpha + \beta {t}^{2}$

How to calculate acceleration

$a = \frac{dv}{dt} \\ v = \alpha + \beta {t}^{2} \\ \frac{dv}{dt} = 0 + 2 \beta t$

if Time is given as 3s

therefore, Acceleration is

$a = 2 \beta t \\ a = 2 \beta 3 \\ a = 2 \times \beta \times 3 \\ a = 6 \beta$

Acceleration is

$a = 6 \beta$

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