# Thread: What is the next step?

1. ## What is the next step?

So the question is:

Suppose an object moves along a horizontal path with the position function
x(t) = 4t t^2 on the time interval 0 ≤ t 5, where x is measured in meters and t is measured in seconds. Find the instantaneous velocity of the object from t = 1. Use units to describe what this value means.

I know that the first step is to find the derivative of the position function so:
lim(delta x -> 0) [s(t + delta t) - s(t)] / [delta t]
[((4t + delta t) - t^2) - (4t - t^2)] / [delta t]

Now I don't understand how to simplify this so that I can plug in 1 for t?

I'm sorry if this looks confusing.
Any help is appreciated.

2. Simply calculate the derivative? The velocity is indeed described by that, so:

$\frac{\partial (4t - t^2)}{\partial t} = 4 - 2t$

If you plug in t = 1 your derivative will be 4 - 2 = 2. The unit of this derivative is meters / second since you derived position (in meters) to time (in seconds).

3. Originally Posted by iluvmathbutitshard
So the question is:

Suppose an object moves along a horizontal path with the position function
x(t) = 4t t^2 on the time interval 0 ≤ t 5, where x is measured in meters and t is measured in seconds. Find the instantaneous velocity of the object from t = 1. Use units to describe what this value means.

I know that the first step is to find the derivative of the position function so:
lim(delta x -> 0) [s(t + delta t) - s(t)] / [delta t]
[((4t + delta t) - t^2) - (4t - t^2)] / [delta t]
This is incorrect. What you want is $\lim_{\Delta t\to 0}\frac{4(t+ \Delta t)- (t+ \Delta t)^2)- 4t+ t^2}{\Delta t}$

Multiply out $4(t+ \Delta t)$ and $(t+ \Delta t)^2$ and cancel what you can. You should be able to cancel everything in the numerator that does NOT have a " $\Delta t$" in it, factor $\Delta t$ out of the numerator and cancel with the $\Delta t$.

Now I don't understand how to simplify this so that I can plug in 1 for t?

I'm sorry if this looks confusing.
Any help is appreciated.