Calculus based physics question??

Question

Okay, well, I was assigned a problem on how to solve for instantaneous velocity and I don't know how to do it. I'm in AP Calculus based Physics.

I'm not asking for answers. I just want to know how to get it. I already have the answers but I need to know how to get to that stage.

It goes like this:

The position of a particle moving along the x-axis is given in centimeters by x=9.75+1.50t^3. Consider the time interval between 2 seconds and 3 seconds and calculate (a) the average velocity, (b) the instantaneous velocity at 2s, (c) the instantaneous velocity at 3s, (d) the instantaneous velocity at 2.5s, and the instananeous velocity when the particle is midway between its positions at t=2s and t=3s.

Would you use a limit function? Or dv/dt?

If you use either, how would you solve it?

Answer 1

Would you use a limit function? Or dv/dt?

dv/dt is the limit of change in velocity when change in time tends to zero.

Since here we have to find the instantaneous velocity we must find the limit of ∆x /∆t when ∆t tends to zero.

In other words we have to find dx / dt.

The given equation x=9.75+1.50t^3 gives the relation between x and t.

If substitute the values for t, t = 2, t= 2.5 and t = 3, the above equation will give us the position x of the particle at various times noted above.

Thus at t = 2, x = 21.75 cm

t = 2.5, x = 33.1875 cm

t = 3, x = 50.25 cm


The average velocity is the total distance traveled from 2 second to 3 second divided by the time interval.


Since we know the distance traveled for 3 s as = 50.25 cm and for 2 s as 21.75 cm
The distance traveled from 2 to 3 s is 50.25 - 21.75 = 28.5 cm

The average speed is 28.5 / [3 - 2] = 28.5 [cm] / s
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To find the instantaneous velocities at various times, we must have an equation relating the instantaneous velocity and time.

To find that equation we find dx/ dt from the equation x=9.75+1.50t^3, dx/dt is the instantaneous velocity at any instant of time t.

dx/dt = 4.5 t^2.

Substitute the values for t and find the instantaneous speeds or velocity at that instant.
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If you use limits you will get the same answer.

Given

x = x=9.75+1.50t^3 After a small interval of time t = t + ∆t

x + ∆x = 9.75+1.50 [t + ∆t] ^3 =

Therefore

∆x = 1.50 {[t + ∆t] ^3 - t^3]

∆x = 1.50 {3 t^2*∆t + 3 t*∆t ^2}

∆x /∆t = 1.50 {3 t^2*+ 3 t*∆t}

When ∆t tends to zero,

dx/dt = 1.50 {3 t^2] = 4.5 t^2.

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Answer 2

For (a), the average velocity = (x(3) - x(2)) / (3-2).
For the rest, the instantaneous velocity is v(t) = dx/dt = 4.50t^2. That immediately gives you (b), (c), and the first half of (d). For the second half of (d), you need to find the position in question (this is (x(2) + x(3)) / 2), then plug it into the equation for x to solve for t. Once you have the value of t, plug it into v(t) to get the answer.