Solve Position at t=3s for Motion Homework: Dynamics Help

[vdfortd]

Homework Statement

An engineer designing a system to control router for a machining process models the system so that the router's acceleration (in in/s²) during an interval of time is given by a= -0.4v, where v is the velocity of the router in in/s². When t=0, the position is s=0 and the velocity is v= 2 in/s. What is the position at t=3 seconds?

Homework Equations

a= dv/dt

The Attempt at a Solution

This section has to do with straight line motion when the acceleration depends on velocity or position.

This is what I tried to do:

a= -0.4v
a= dv/dt

dv/dt = -0.4v

dv/v= -0.4 dt

I integrated both sides and I got:

ln v = -.4t + C ( C = constant)

From my differential equations math class, we did problems like these and we got rid of the ln by making it all a power of e.

e^{ln v} = e ^{-.4t + C}

you get v= e^-.4t+C or v= e^-.4tC

Now I know this is wrong because we know that when t=0 v=2. Am I on the right track or did i do this completely wrong? Thanks for any help.

 

[nvn]

vdfortd: Nice work. You are doing well, so far. You said, "Now I know this is wrong." That is not true. Go ahead and substitute t = 0 and v = 2 into your last equation, and see what you get. Solve for C. Now continue solving the problem.

 

[Mene]

You are on the right track, but there are a few errors in your approach. Let's start with the differential equations:

a = dv/dt

We can rewrite this as:

dv = a dt

Now, we can integrate both sides to get:

∫dv = ∫a dt

v = at + C

Now, we need to solve for the constant C. We can use the initial conditions given in the problem to do this. When t=0, v=2, so we can substitute these values into our equation:

2 = a(0) + C

C = 2

Now, we have our final equation for velocity:

v = at + 2

Next, we can use this equation to solve for position. From the homework statement, we know that position is defined as the integral of velocity with respect to time:

s = ∫v dt

Substituting our equation for velocity, we get:

s = ∫(at + 2) dt

s = (1/2)at^2 + 2t + C

Again, we can use the initial conditions to solve for the constant C. When t=0, s=0, so we have:

0 = (1/2)a(0)^2 + 2(0) + C

C = 0

Now, we have our final equation for position:

s = (1/2)at^2 + 2t

To find the position at t=3 seconds, we can simply substitute t=3 into our equation:

s = (1/2)a(3)^2 + 2(3)

s = (1/2)(-0.4)(3)^2 + 2(3)

s = -2.7 inches

Therefore, the position at t=3 seconds is -2.7 inches.

 

[Mene]

Your approach is correct so far. However, you have made a small mistake when integrating. The correct solution should be v = Ce^(-0.4t), where C is the constant of integration.

To solve for the constant C, we can use the initial conditions given in the problem. When t=0, v=2, so we can substitute these values into the equation:

2 = Ce^(-0.4(0))

2 = C

Therefore, the equation for velocity becomes v = 2e^(-0.4t).

To solve for position, we can integrate the velocity equation:

s = ∫v dt = ∫2e^(-0.4t) dt = -5e^(-0.4t) + D

Again, we can use the initial condition s=0 at t=0 to solve for the constant D:

0 = -5e^(-0.4(0)) + D

0 = -5 + D

D = 5

Therefore, the final equation for position is s = -5e^(-0.4t) + 5.

To find the position at t=3 seconds, we can simply plug in t=3 into the equation:

s(3) = -5e^(-0.4(3)) + 5 = 0.196 inches.

So, the position at t=3 seconds is 0.196 inches.