How Do You Calculate Instantaneous Velocity at a Midpoint in Particle Motion?

[fight_club_alum]

Homework Statement

The position of a particle moving along the x axis is given in centimeters by x = 9.35 + 1.03 t3, where t is in seconds. Calculate the instantaneous velocity when the particle is midway between its positions at t = 2.00 s and t = 3.00 s.

Homework Equations

The Attempt at a Solution

f(2)-f(3)/2
then, solve for t, which will be cubic_root(17.5)
add it to the equation (f(to+h)-f(to))/h as h tends to 0

 

[kuruman]

Once you have t, you have to put that value in the equation for v(t). What is that equation and what did you get for the value of t?

 

[fight_club_alum]

Once you have t, you have to put that value in the equation for v(t). What is that equation and what did you get for the value of t?

Once you have t, you have to put that value in the equation for v(t). What is that equation and what did you get for the value of t?

8.022

 

[fight_club_alum]

Once you have t, you have to put that value in the equation for v(t). What is that equation and what did you get for the value of t?

I use this equation for the evaluation of tangent, instantaneously, which is in this case is the instantaneous velocity lim h -->0 f(to+h) - f(to) / h

 

[kuruman]

Since you know $x(t)$, It is better to use $$v(t)=\frac{dx(t)}{dt}.$$

On edit:

8.022

I disagree with this value. Can you show how you got it?

 

[fight_club_alum]

Since you know $x(t)$, It is better to use $$v(t)=\frac{dx(t)}{dt}.$$

On edit:

I disagree with this value. Can you show how you got it?

I followed many steps that would be hard to post here

 

[Ray Vickson]

I followed many steps that would be hard to post here

How can we possibly help if you will not show what you have done? If you know how to take derivatives, it is simple---hardly needing any work. If you don't know how to take derivatives you will need to a bit more work, but it is still manageable.

Just be careful how you write things: when you write f(t0+h) - f(t0), the standard parsing rules for mathematical expressions interprets this as
$$ f(t_0+h) - \frac{f(t_0)}{h},$$
which (I hope) you do not intend. If you mean
$$ \frac{f(t_0 +h) - f(t_0)}{h}$$
you need to use parentheses, like this: [f(t0+h)-f(t0)]/h. That forces the subtraction in the numerator to occur before the division.

 

[Chestermiller]

What is the value of f(3)? What is the value of f(2)? What is the location half-way between these two locations? What is the time corresponding to this half-way location?