Question

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?.

Answer 1

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

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

At the endpoints of the the interval, we have

`v(2.0\,\mathrm s) = 2.3(\rm m)/(\rm s)+\left(4.1(\rm m)/(\mathrm s^2)\right)(2.0\,\mathrm s)-\left(6.2(\rm m)/(\mathrm s^3)\right)(2.0\,\mathrm s)^2 = -14.3(\rm m)/(\rm s)`

`v(1.0\,\mathrm s) = 2.3(\rm m)/(\rm s)+\left(4.1(\rm m)/(\mathrm s^2)\right)(1.0\,\mathrm s)-\left(6.2(\rm m)/(\mathrm s^3)\right)(1.0\,\mathrm s)^2 = 0.2(\rm m)/(\rm s)`

Then the average acceleration is

`a_{\mathrm{ave}[1.0\,\mathrm s,2.0\,\mathrm s]} = (-14.3(\rm m)/(\rm s)-0.2(\rm m)/(\rm s))/(1.0\,\mathrm s) = \boxed{-14.5(\rm m)/(\mathrm s^2)}`