# Instantaneous velocity

• Sep 24th 2008, 01:43 PM
silencecloak
Instantaneous velocity
Man, these types of problems are given me a run for my money

$\displaystyle t2 - 7t + 16$

Find the instantaneous velocity at t = 4

I get this problem simplified down to

$\displaystyle \frac{x(x-7)-a(a-7)}{x-a}$

Which I know is right.

But can i cancel the denominator by making the equation $\displaystyle \frac{(x-7)(a-7)}{1}$ then plugging in 4 and getting 6
• Sep 24th 2008, 01:52 PM
mr fantastic
Quote:

Originally Posted by silencecloak
Man, these types of problems are given me a run for my money

$\displaystyle t2 - 7t + 16$

Find the instantaneous velocity at t = 4

I get this problem simplified down to

$\displaystyle \frac{x(x-7)-a(a-7)}{x-a}$

Which I know is right.

But can i cancel the denominator by making the equation $\displaystyle \frac{(x-7)(a-7)}{1}$ then plugging in 4 and getting 6

What's $\displaystyle t^2 - 7t + 16$ meant to be? Displacement? Acceleration? Velocity? Without knowing this, your questions can't be answered.
• Sep 24th 2008, 01:53 PM
silencecloak
Quote:

Originally Posted by mr fantastic
What's $\displaystyle t^2 - 7t + 16$ meant to be? Displacement? Acceleration? Velocity? Without knowing this, your questions can't be answered.

"The displacement (in meters) of a particle moving in a straight line is given by s = t2 - 7t + 16, where t is measured in seconds."
• Sep 24th 2008, 01:55 PM
mr fantastic
Quote:

Originally Posted by silencecloak
"The displacement (in meters) of a particle moving in a straight line is given by s = t2 - 7t + 16, where t is measured in seconds."

v = ds/dt. Then substitute t = 4.
• Sep 24th 2008, 01:57 PM
silencecloak
Quote:

Originally Posted by mr fantastic
v = ds/dt. Then substitute t = 4.

that just confuses me :(

I have to use the definition

$\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

which is how i got down to the equation i posted above
• Sep 24th 2008, 02:13 PM
mr fantastic
Quote:

Originally Posted by silencecloak
that just confuses me :(

I have to use the definition

$\displaystyle \lim_{x \to a} \frac{f(x)-f(a)}{x-a}$

which is how i got down to the equation i posted above

1. Drip feeding important information over a series of posts in a thread doesn't help your cause in getting fast, simple and effective help.

2. Using symbols like x and a in a motion problem is just plain confusing when they do not represent displacement and acceleration.

You need to develop understanding on how to use these formulae with notation appropriate to the context. You also need to understand why this is important.

$\displaystyle v(4) = \lim_{t \rightarrow 4} \frac{(t^2 - 7t + 16) - (4^2 - 7(4) + 16)}{t-4} = \lim_{t \rightarrow 4} \frac{t^2 - 7t - 12}{t-4}$ $\displaystyle = \lim_{t \rightarrow 4} \frac{(t - 4)(t - 3)}{t-4} = \lim_{t \rightarrow 4} (t - 3) = 1$.
• Sep 24th 2008, 02:19 PM
silencecloak
Quote:

Originally Posted by mr fantastic
1. Drip feeding important information over a series of posts in a thread doesn't help your cause in getting fast, simple and effective help.

2. Using symbols like x and a in a motion problem is just plain confusing when they do not represent displacement and acceleration.

You need to develop understanding on how to use these formulae with notation appropriate to the context. You also need to understand why this is important.

$\displaystyle v(4) = \lim_{t \rightarrow 4} \frac{(t^2 - 7t + 16) - (4^2 - 7(4) + 16)}{t-4} = \lim_{t \rightarrow 4} \frac{t^2 - 7t - 12}{t-4}$ $\displaystyle = \lim_{t \rightarrow 4} \frac{(t - 4)(t - 3)}{t-4} = \lim_{t \rightarrow 4} (t - 3) = 1$.

Sorry I thought it was implied, and I am just now learning this in school..so...sorry...
• Sep 24th 2008, 02:27 PM
hockey777
isn't instantaneous velocity acceleration, so he has to take the derivative twice?
• Sep 24th 2008, 02:37 PM
mr fantastic
Quote:

Originally Posted by hockey777
isn't instantaneous velocity acceleration, so he has to take the derivative twice?

NO!