Question

An object moves along the x axis according to the equation x(t) = (3.00 t^2 - 2.00t + 3.00) m, where t is in seconds.
Determine: (a) the average speed between t= 2.00 s and t= 3.00 s,
(b) the instantaneous speed at t= 2.00 s and at t= 3.00 s,
(c) the average acceleration between t= 2.00 s and t= 3.00 s, and (d) the instantaneous acceleration at t= 2.00 s and t= 3.00 s. ( i cant get why u solved part c from part b not a

Answer 1

Alright, speed is defined as:
\[\left| \vec{v} \right|=\frac{\Delta x}{\Delta t}=\frac{x_f-x_i}{t_f-t_i}\]Where, specifically, the instantaneous speed is defined as:\[\lim_{\Delta \rightarrow 0}\frac{\Delta x}{\Delta t}\]Can you answer a and b from that information?

Answer 2

Whoops. That's a typo. Should be:
\[\left| \vec{v_{inst}} \right|=\lim_{\Delta t \rightarrow 0}\frac{\Delta x}{\Delta t}\]

Answer 3

i know how to solve part a
but in part b i dont get why we take the derivative
and in part c i dont get why we use part b

Answer 4

A derivative of a function is the rate of change with respect to the variable of differentation.

Since x(t) represents the position of the particle at all time, t, by taking its derivative (with respect to t), you will get the rate of change of position (otherwise known as velocity) of the particle at all time, t.

The instantaneous velocity is similar to the average velocity except the portion of time you are looking over is infinitesimally small.

The reason you'll use the answers from part b for part c is simply that the definition of acceleration is:
\[\vec{a}=\frac{\Delta \vec{v}}{\Delta t}\]And of course, in the limit:
\[\lim_{\Delta t \rightarrow 0}\vec{a}=\frac{d\vec{v}}{dt}\]and so the acceleration is the first time derivative of the velocity.

Since we know that the velocity is the first time derivative of position, we can infer that:\[\vec{a}=\frac{d\vec{v}}{dt}=\frac{d^2\vec{x}}{dt^2}\]

Velocity is the rate of change of position. Acceleration is the rate of change of velocity.