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The position of an object is given by x = at3 - bt2 + ct,where a = 4.1 m/s3, b = 2.2 m/s2, c = 1.7 m/s, and x and t are in SI units. What is the instantaneous acceleration of the object when t = 2.8 s?

Respuesta :

Answer:

64.48 m/s^2

Explanation:

The computation of the instantaneous acceleration of the object is as follows:

Given that

x = at3 - bt2 + ct

Now Instantaneous velocity,

v = dx ÷ xt

= 3at2 - 2bt + c

And, Instantaneous acceleration, A is

= dv ÷ dt

= 6at - 2b

Now put the value of a, b and t in the above equation

A = 6 × 4.1 × 2.8 - 2 × 2.2

= 64.48 m/s^2

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