How to Find Instantaneous Velocity at a Specific Time?

[[Nicolette]]

Homework Statement

Find the instantaneous velocity at t=1 by computing v(1)= [h(1+$$\Delta$$t)-h(1)]/$$\Delta$$t

I found that v(1)= -8ft/sec. Also I know h(1)=24 but i don't understand how to manipulate the h(1+$$\Delta$$t) to get the solution.

Homework Equations

h(t)=16+24t-16t²

The Attempt at a Solution

The solution the teacher gave is v(1)=-8-16t$$\Delta$$

 

[rootX]

What's h(1+dt)? Substitute (1+dt) into the h(t) equation.. and put everything in the v(1)= [h(1+dt)-h(1)]/dt equation. If you expand the equation and do some algebra you will eventually get the answer

 

[nlightner]

t.

To find the instantaneous velocity at t=1, we need to use the definition of instantaneous velocity, which is the derivative of position with respect to time. In this case, h(t) represents the position of an object at time t, so we can use the derivative of h(t) to find the instantaneous velocity at t=1.

The derivative of h(t) is given by h'(t)=24-32t. We can plug in t=1 to find the instantaneous velocity at t=1:

h'(1) = 24-32(1) = -8 ft/sec

This matches the value of -8 ft/sec that you found earlier. So, the solution given by your teacher is correct.

To better understand how to manipulate h(1+\Deltat), we can use the formula for h(t) given in the homework statement:

h(t)=16+24t-16t^2

To find h(1+\Deltat), we simply plug in t=1+\Deltat to the formula:

h(1+\Deltat)=16+24(1+\Deltat)-16(1+\Deltat)^2

=16+24+24\Deltat-16-32\Deltat-16\Deltat^2

Simplifying, we get:

h(1+\Deltat) = 24+8\Deltat-16\Deltat^2

Now, we can use this to find the value of v(1) by plugging in these values into the formula given in the homework statement:

v(1)=[h(1+\Deltat)-h(1)]/\Deltat

= [24+8\Deltat-16\Deltat^2-24]/\Deltat

= [8\Deltat-16\Deltat^2]/\Deltat

= 8-16\Deltat

= -8-16\Deltat

= -8-16(0) (since we are finding the instantaneous velocity at t=1, so \Deltat=0)

= -8 ft/sec

Therefore, we can see that the solution given by your teacher is correct. It's important to understand that h(1+\Deltat) represents the position of the object at a time slightly after t=1, and by subtracting h(1) from it