Acceleration, Velocity and Position (calculus)

In summary, the conversation discusses finding the velocity, acceleration, and position of a particle at a given time and position, using the equation v = kt + C. After solving for C and finding the value of k, the correct equation for v is v(t) = 16 - t^2. With this equation, it is possible to solve for the position when t = 7 and verify the correct answer.
  • #1
adoado
72
0

Homework Statement



The acceleration of a particle is directly proportional to time, t. At t = 0, the velocity of that particle, v, = 16 in./sec. Knowing that v = 15 in/sec and x = 20in when t = 1, determine the velocity, acceleration and position at x = 7.

Homework Equations



The Attempt at a Solution



I took
[tex]\frac{dv}{dt} = kt, so \: v = \frac{k}{2}t^2 + C[/tex]

Solving for C:
[tex]16 = \frac{k}{2}(0)^2 + C[/tex]
C = 16

Solve for k

[tex]1 = \frac{k}{2}(1)^2 + 16[/tex]

k = -30

Therefore:

[tex]v(t) = \-15*(t)^2 + 16[/tex]

This is already wrong; v(7) = -719 in/sec but the book states the answer is -33 in/sec

Any advice on what's gone wrong would be greatly appreciated; this is a revision question too...

Cheers,
Adrian
 
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  • #2
k is -2
 
  • #3
You need to find the velocity when x = 7. Find x(t) first.

ehild
 
  • #4
Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?
 
  • #5
adoado said:
Ah, thankyou ehild; I failed to read the question once again... I will give it a shot. How is k = -2?

Looks OK.
 
  • #6
I found x(t) by integrating and finding C. I obtained this:

[tex]x(t)=16t-5t^3+9[/tex]

Then solving for t when x = 7

[tex]16t-5t^3+2 = 0[/tex]

Sorry, how does one solve this?

Cheers,
Adrian
 
Last edited:
  • #7
Don't forget that k=-2, so v(t)=16-t^2.

Looking at the solution for v(t) at x=7 , it should be rather t=7 instead of x. Otherwise why would they ask the position at x=7?

ehild
 
Last edited:
  • #8
Sorry, this is probably really stupid, but where/how does k = -2?
 
  • #9
"v = 15 in/sec and x = 20in when t = 1"

15=k/2*1^2+16

ehild
 
  • #10
Ah! I put in a bad number...dodgy mistake indeed; Cheers, I will retry!
 
  • #11
Got it! Thanks so much, both of you! ;)
 

1. What is the difference between acceleration, velocity, and position?

Acceleration is the rate of change of velocity over time, velocity is the rate of change of position over time, and position is the location of an object in space relative to a reference point.

2. How are acceleration, velocity, and position related in calculus?

In calculus, acceleration is the first derivative of velocity, and velocity is the first derivative of position. This means that acceleration is the second derivative of position.

3. How do you calculate acceleration, velocity, and position using calculus?

To calculate acceleration, velocity, and position using calculus, you would need to take derivatives and integrals of the given functions. The derivative of position gives velocity, and the derivative of velocity gives acceleration. To find position, you would need to integrate the function for velocity, and to find velocity, you would need to integrate the function for acceleration.

4. How does acceleration affect an object's motion?

Acceleration affects an object's motion by changing its velocity. If an object is accelerating, it is either speeding up or slowing down. The direction of the acceleration also determines the direction of the change in velocity.

5. What are some real-life applications of calculus in understanding acceleration, velocity, and position?

Calculus is used in many real-life applications to understand acceleration, velocity, and position. For example, it is used in physics to study the motion of objects, in engineering to design and optimize structures and machines, and in economics to model and analyze financial data. Calculus is also used in fields such as astronomy, biology, and chemistry for various applications related to acceleration, velocity, and position.

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