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Solving without using differentiation?

  • #1

Homework Statement




Hi all, I'm trying to understand why my second way of solving this problem doesn't work:

A 3.00kg object is moving in a plane, with its x and y coordinates given by x= 5t^2–1 and y=3t^3 + 2, where x and y are in meters and t is in seconds. Find the magnitude of the net force acting on this object at t = 2.00s


Homework Equations



xf = xi + vi(t) + (1/2)a(t^2)



The Attempt at a Solution



I've found the answer with differentiation but I don't think we are supposed to use differentiation in my class so I'm wondering why when I plug the numbers into one of the kinematic equations, I get the wrong answer? I evaluate the object's position at t = 2.00s to be (19 i + 26 j) m and now since I have two components of the objects final position:

19 m = (1/2)ax t^2
26 m = (1/2)ay t^2

to get the acceleration components in order to solve for the Force but this yields the wrong answer. Any clarification is appreciated.
 

Answers and Replies

  • #2
23
1
It looks like you're missing something in your bottom equations... What happened to your initial conditions?
 
  • #3
It looks like you're missing something in your bottom equations... What happened to your initial conditions?
I guess I assumed that the initial velocity was zero because plugging t=0 into the derivative of the equations would get you 0 from the velocities? I used differentiation to come to that conclusion.

Oh wait a second, I guess the initial positions wouldn't be zero since plugging t = 0 would result in x-component of -1 and y-component of +2. I wonder if this would result in the right answer. I will try
 
  • #4
23
1
Good luck! It should work out.
 
  • #5
Well I plugged in my new values for initial position and got 46.9 N but I don't think that's the right answer.

19 m = -1 m + 0 + (1/2)ax(t^2)

ax = 10 m/s^2

26 m = 2 m + 0 + (1/2)ay(t^2)

ay = 12 m/s^2

so Fx = (3.00 kg)(10 m/s^2) = 30 N
and Fy = (3.00 kg)(12 m/s^2) = 36 N

Ft = 46.9 N

I'm pretty sure the answer is supposed to be around 112 N so I guess I'm missing something else? Hmm
 
  • #6
gneill
Mentor
20,793
2,773
You're going to have trouble casting the y-component equation into the form

d = do + vo*t + (1/2) a*t2

because it has a dependency on t3.

When equations of motion have a t3 term, it's called a Jerk, and the form of the general equation becomes:

d = do + vo*t + (1/2) a*t2 + (1/6) J*t3

In such a case, the acceleration at time t becomes

a + J*t

So, cast your y-component equation into this general form and go at it.
 
  • #7
cepheid
Staff Emeritus
Science Advisor
Gold Member
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The kinematic equations you are trying to use are ONLY valid for motion under constant accleration. To see that that is true, take, take x = x0 + v0 + 1/2at^2 and differentiate it twice. Or, take a(t) = a = const. and integrate it twice to get back what you started with.

You see the inconsistency, right? x(t) and y(t) are either equal to this expression, OR they are equal to what was given in the problem. But not both.

In the x case, it still works, since the given x(t) is actually of this form -- it varies quadratically with t, and hence the x-acceleration is indeed constant.

In the y-case it doesn't work. y(t) varies cubically with t, which is not consistent with constant acceleration. Differentiate it twice and see what you get. You'll find that the y-acceleration isn't constant.
 
Last edited:
  • #8
Wow ok thank you for that information about cubic degrees and their relation to constant acceleration, I will stick to my derivative answer but I was just curious as to other types of solutions because I know that many students in the class have not completed their differential calculus seeing as it is only a co-requisite to the Physics we're taking. Cheers!
 
  • #9
gneill
Mentor
20,793
2,773
Well, just for fun, let's see what happens when we do it via the general form of the equation of motion that includes jerk.

You gave us:

x= 5t^2–1 and y=3t^3 + 2

which, cast into the general form, become the independent equations:

x(t) = (1/2)(10m/s2)t2 - 1m

y(t) = (1/6)(18m/s3)t3 + 2m

The acceleration at time t is then:

X-component: 10m/s2

Y-component: 0 + (18m/s3)t

So at time t = 2.00 seconds the net acceleration becomes,

[tex] a = \sqrt{10^2 + (18*2.00)^2} m/s^2 [/tex]

[tex] a = 37.363 m/s^2 [/tex]

and the net force is then

[tex] a*3.00 kg = 112 N [/tex]
 
  • #10
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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..but I was just curious as to other types of solutions because I know that many students in the class have not completed their differential calculus seeing as it is only a co-requisite to the Physics we're taking. Cheers!
It is preferred to take calculus as a prerequisite prior to taking calculus based physics, but sometimes they are taken concurrently, and it places the student at a disadvantage when trying to understand the derivations of formulae or doing problems with non constant acceleration and forces, etc. Fortunately, most problems in Intro Physics get into the calculus mostly only for derivations or very basic problem solving.
 

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