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?

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

force is mass*acceleration.

we have a very small mass: 11.4*0.001 kg

if we want 0.035n of force, we need an acceleration:

0.035 = 11.4*0.001*a

a = 0.035/(11.4*0.001) m/s^2