Dynamics Question using nonconstant acceleration

In summary, a particle with an initial speed of 27 m/s experiences a deceleration of a = -(6t) m/s^2. Using the equation a = dv/dt and integrating, the time it takes for the particle to stop is 3 seconds. To find the distance traveled, the equation v = ds/dt is used and integrated, resulting in s = (-3/3)*t^3+v0*t+s0. Substituting t = 3 seconds, the distance traveled before the particle stops is 40.5 meters.
  • #1
gatrhumpy
3
0

Homework Statement


A particle has an initial speed of 27 m/s. If it experiences a deceleration of a = -(6t) m/s^2, where t is in seconds, determine the distance traveled before it stops.


Homework Equations



a = dv/dt
v = ds/dt
ads = vdv (not independent from the above two equations)

The Attempt at a Solution



What I know: v(0) = initial speed = 27 m/s
v = final speed = 0 m/s
t(0) = initial time = 0 seconds (assumption)
s(0) = initial displacement = 0 meters (assumption)

I used a = dv/dt and integrated to find the time for the particle to stop. I found it this way:

a = dv/dt = (-6t). dv = (-6t)dt. Lower limit for v(0) = 27, upper limit for v = 0. Lower limit for t(0) = 0, upper limit for t = t. Integrating dv = (-6t)dt, I get -v(0) = -3t^2 --> t = 3 seconds. This is the time it takes the particle to stop.

However it asks for the distance to stop, and I have no idea how to get it. I know I can't use constant acceleration formulas because the acceleration is a function of time. I tried using ads = vdv as that is independent of time, but the acceleration a is dependent on time as a = (-6t) m/s^2
 
Last edited:
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  • #2
In your problem statement you state a constant decelaration of 6 m/s^2
Then you go on with: a = dv/dt = (-6t), which is the correct defintion
Integration produces : v = -3*t^2+v0. You got the integration idea very right.
t = 3 sec.
So, continue with: v = ds/dt in a similar way.
v = ds/dt integration gives: s = (-3/3)*t^3+v0*t+s0
You can solve this for t= 3 sec.
 
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  • #3
Thanks man.

I'm 're-learning' dynamics for my FIT exam in October. Funny how the simplest solution is right there in front of you and yet, nothing. :)
 
  • #4
Why din you try 2nd Order Differential Equation? differentiate twice will induce many unnecessary constants.
 
  • #5
darkdream said:
Why din you try 2nd Order Differential Equation? differentiate twice will induce many unnecessary constants.

Probably could have, but haven't reviewed DEs yet. :)
 

1. What is nonconstant acceleration?

Nonconstant acceleration is when an object's acceleration changes over time. This means that the object's speed is changing at different rates, either increasing or decreasing.

2. How is nonconstant acceleration different from constant acceleration?

Constant acceleration is when an object's acceleration remains the same throughout its motion, while nonconstant acceleration means that the acceleration is changing.

3. How is nonconstant acceleration calculated?

Nonconstant acceleration can be calculated using the formula a = (vf - vi)/t, where a is the acceleration, vf is the final velocity, vi is the initial velocity, and t is the time interval.

4. What are some real-life examples of nonconstant acceleration?

Some examples of nonconstant acceleration in everyday life include a car accelerating from a stop, a roller coaster going up and down a hill, or a pendulum swinging back and forth.

5. How does nonconstant acceleration affect an object's motion?

Nonconstant acceleration can affect an object's motion by causing it to speed up or slow down, or by changing its direction. It can also cause the object to have a curved rather than straight path.

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