contestada

The velocity v(t) of a particle as a function of time is given by v(t) = (2. 3 m/s) + (4. 1 m/s2)t - (6. 2 m/s3)t2. What is the average acceleration of the particle between t = 1. 0 s and t = 2. 0 s?.

Respuesta :

By definition of average acceleration, the ave. acc. for this particle between 1.0 s and 2.0 s is

[tex]a_{\mathrm{ave}[1.0\,\mathrm s,2.0\,\mathrm s]} = \dfrac{v(2.0\,\mathrm s) - v(1.0\,\mathrm s)}{2.0\,\mathrm s - 1.0\,\mathrm s}[/tex]

At the endpoints of the the interval, we have

[tex]v(2.0\,\mathrm s) = 2.3\dfrac{\rm m}{\rm s}+\left(4.1\dfrac{\rm m}{\mathrm s^2}\right)(2.0\,\mathrm s)-\left(6.2\dfrac{\rm m}{\mathrm s^3}\right)(2.0\,\mathrm s)^2 = -14.3\dfrac{\rm m}{\rm s}[/tex]

[tex]v(1.0\,\mathrm s) = 2.3\dfrac{\rm m}{\rm s}+\left(4.1\dfrac{\rm m}{\mathrm s^2}\right)(1.0\,\mathrm s)-\left(6.2\dfrac{\rm m}{\mathrm s^3}\right)(1.0\,\mathrm s)^2 = 0.2\dfrac{\rm m}{\rm s}[/tex]

Then the average acceleration is

[tex]a_{\mathrm{ave}[1.0\,\mathrm s,2.0\,\mathrm s]} = \dfrac{-14.3\frac{\rm m}{\rm s}-0.2\frac{\rm m}{\rm s}}{1.0\,\mathrm s} = \boxed{-14.5\dfrac{\rm m}{\mathrm s^2}}[/tex]