Solve Calc-Based Kinetics Problem: Initial Velocity for Washington Monument

[lLovePhysics]

Homework Statement

With what intial velocity must an object be thrown upward (from ground level) to reach the top of the Washington Monument (approx 550ft)?

Homework Equations

Here's what I know:

s(0)=0
a(t)=-32ft/s^2
$$s(t_{max})=550\\<br /> s'(t_{max})=0$$

The Attempt at a Solution

Actually I got the correct answer but I don't understand something. How do you know whether the constants "C" given by the indefinite integrals are the same?

For example, when you integrate a(t)=s''(t) you get:

s'(t)=-32+C

When you integrate the velocity or s'(t) you get:

$$s(t)=-16t^2+Ct+C$$

So are those C's the same or are they different? How do you know? When I treated them the same I got the correct answer, but when I didn't the answer turned out to be wrong.

Can someone please explain the "constants dilemma?" Thanks.

 

[Dick]

They are different. And s'(t)=-32t+C, you left out the t. You fix the constants by making sure that s(0)=0 and the max of s(t) is 550ft.

 

[dynamicsolo]

For example, when you integrate a(t)=s''(t) you get:

s'(t)=-32t+C [I corrected this line.]

Since you integrated the (constant) acceleration function, you now have the velocity function. The velocity at time t = 0 would be

v(0) = s'(0) = -32·0 + C = C ,

so the "arbitrary constant" becomes your initial velocity v(0). Now,

When you integrate the velocity or s'(t) you get:

$$s(t) = -16t^2 + Ct + D$$ [it's a good idea to call the second arbitary constant something else]

The position function is now

s(t) = -16(t^2) + v(0)·t + D ,

which at time t = 0 becomes

s(0) = -16·(0^2) + v(0)·0 + D .

So the second arbitrary constant is D = s(0), the initial position. This is where the textbooks get the formulas for constant acceleration (a) kinematics

v(t) = v(0) + at ,

x(t) = x(0) + v(0)·t + (1/2)·a·(t^2) .

 

[lLovePhysics]

WOW That's so cool! Okay thanks guys. First time seeing things in both the calculus and physics perspectives. lol

 

[dynamicsolo]

WOW That's so cool! Okay thanks guys. First time seeing things in both the calculus and physics perspectives. lol

I really wish the two were simply taught together, since they grew up together. A great deal of mathematical technique and theory for millenia, but particularly over the last four hundred years, was in aid of solving increasingly sophisticated problems in physics and engineering...

 

[lLovePhysics]

I really wish the two were simply taught together, since they grew up together. A great deal of mathematical technique and theory for millenia, but particularly over the last four hundred years, was in aid of solving increasingly sophisticated problems in physics and engineering...

Yeah, me too. Unfortunately I learned physics before learning calculus, which I'm currently taking. I wish I had learned both together. I'm glad I took physics though, or else I wouldn't understand the physics-based calculus problems as much.

 

[lLovePhysics]

Also, I can't believe I never knew how to actually derive these formulas.. I guess I couldn't anyways since I did not know any calculus to begin with.

 

[dynamicsolo]

Also, I can't believe I never knew how to actually derive these formulas.. I guess I couldn't anyways since I did not know any calculus to begin with.

Actually, the constant acceleration kinematic equations were already known in medieval Europe. If you make a graph of a constant acceleration function and ask how the area under it increases as time on the graph advances, you get the velocity equation given above. If you do the same with that function, you get the position equation. (They were doing what we would now called "graphical integration".) Newton developed calculus in order to grapple with problems involving non-constant accelerations.