Acceleration, Velocity and distance

In summary: Because you never used v=2.4 in your integrations. Without that information, you can't actually solve for v.
  • #1
Oblivion77
113
0

Homework Statement



An engineer designing a system to control a router for a machining process models the system so that the router's acceleration during an interval of time is a=-1.8vm/s . When t=0 , its position is s=0 and its velocity is 2.4 m/s. Determine the router's position (unit: m) at s.

Homework Equations



a=dv/dt, v=ds/dt

The Attempt at a Solution



First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!
 
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  • #2
Oblivion77 said:
First i did -1.8v=dv/dt did some integration and work and found v = e^(-1.8t)

I then used e^(-1.8t) = ds/dt and solved for s and subsituted t=0(lower bound) and t=0.75(upper bound) and found s = 0.412m

I am not sure if the first step I did is allowed since I never used the v=2.4. Please tell me my mistakes. Thanks!

Hi Oblivion77! :smile:

You integrated twice, but you only put in a constant of integration the second time.

The v = 2.4 comes in the first time! :wink:
 
  • #3
Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
t=(-1/1.8)ln(v)+0.49 (does this look right so far?). Now I want to solve for v and use v=ds/dt to get the function for position. so v=e^(0.88-1.8t), e^(0.88-1.8t)=ds/dt .Then to sub t=0.75s for the upper bound and t=0 for the lower bound. Does this look right now? Or did I make a mistake somewhere? Thanks!
 
  • #4
Oblivion77 said:
Ok, using the v=2.4 with the first integration I get -1.8v=dv/dt, doing some integration I get
t=(-1/1.8)ln(v)+0.49

oooh :cry:

No, you integrated dv/v - -1.8dt to get v = Ce-1.8t.

And so, when t= 0, e-1.8t = … ? , and so C = … ? :smile:
 
  • #5
Ok, so when i do that I get v=Ce^(-1.8t). And the intial conditions are t=0, v=2.4. So would i do 2.4=Ce^0, C=2.4. Then v=2.4e^(-1.8t), or do i not use the velocity?
 
  • #6
What a long question! :wink:

Yes, v = 2.4e-1.8t

why does that worry you? :smile:
 

FAQ: Acceleration, Velocity and distance

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It measures how quickly an object's speed or direction is changing.

How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. The unit for acceleration is meters per second squared (m/s^2).

What is the difference between acceleration and velocity?

Velocity is the rate of change of an object's displacement, while acceleration is the rate of change of an object's velocity. In simpler terms, velocity tells us how fast an object is moving and in what direction, while acceleration tells us how quickly the object is changing its speed or direction.

How can acceleration affect an object's motion?

Acceleration can affect an object's motion in various ways. If an object is accelerating in the same direction as its velocity, it will speed up. If it is accelerating in the opposite direction, it will slow down. If the acceleration is perpendicular to the velocity, the object will change direction without changing speed. If there is no acceleration, the object will continue to move at a constant speed in a straight line.

What is the relationship between distance, velocity, and acceleration?

Distance, velocity, and acceleration are all related through the fundamental equations of motion. Distance is the product of average velocity and time, while velocity is the product of acceleration and time. This means that if acceleration remains constant, the distance an object travels is directly proportional to the square of the time it takes. In other words, the farther an object travels, the longer it takes to reach that distance.

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