Question

The position of an object at time t is given by s(t) = 6 - 14t. Find the instantaneous velocity at t = 6 by finding the derivative.
By using the difference quotient

Answer 1

Answer:

It results -14 in either way

Step-by-step explanation:

Velocity As A Rate Of Change

The velocity of an object can be computed as the rate of change of its displacement (or position taken as a vector) over time. If we compute it as a derivative, it's called instantaneous velocity, and if computed as the slope of the function (difference quotient) at a certain point it's the average velocity

The position of the object as a function of time is

$\displaystyle s(t)=6-14t$

Computing the derivative

$\displaystyle s'(t)=-14$

We can see it's a constant value. If we use the slope or rate of change:

$\displaystyle v=\frac{s_2-s_1}{t_2-t_1}$

Now let's fix two values for time  

$\displaystyle t_1=5\ sec,\ t_2=8\ sec$

and compute the corresponding positions, by using the given function

$\displaystyle s_1=6-14(5)=-64$

$\displaystyle s_2=6-14(8)=-106$

Now we compute the average velocity

$\displaystyle v=\frac{-106-(-64)}{8-5}$

$\displaystyle v=\frac{-106+64}{3}=\frac{-42}{3}$

$\displaystyle v=-14$

We get the very same result in both ways to compute v. It happens because the position is related with time as a linear function, it's called a constant velocity motion.