Kinematics and acceleration problem

[aaronfue]

Homework Statement

A particle moves along a straight line such that its acceleration is a = (4t²-2) \frac{m}{s^2}, where t is in seconds. When t = 0, the particle is located 1 m to the left of the origin, and when t = 2 s, it is 20 m to the left of the origin. Determine the position of the particle when t = 4.3s.

2. Homework Equations & The attempt at a solution.

I am just stuck on this. I know I should already know how to do this but right now my mind is blank!
I know that I can get the velocity equation by integrating, but when I try to use t=2 s, and then using the equation x=x_o+v_ot+\frac{1}{2}at², I am not even close to the x=20.

 

[tms]

The kinematics equation you mention is good only for constant acceleration, but the acceleration in the problem varies with time, so you have to go back to basics. Start with the definition of acceleration: a = dv/dt.

 

[haruspex]

x=x_o+v_ot+\frac{1}{2}at²

That equation is only valid for constant acceleration.

What do you get when you integrate the acceleration equation?

 

[aaronfue]

That equation is only valid for constant acceleration.

What do you get when you integrate the acceleration equation?

After I integrate the acceleration, I get:

v(t)= (\frac{4}{3}t³ - 2t) \frac{m}{s}

 

[haruspex]

After I integrate the acceleration, I get:

v(t)= (\frac{4}{3}t³ - 2t) \frac{m}{s}

Don't forget the constant of integration.
Then get the equation for distance as a function of time.

 

[tms]

Don't forget the constant of integration.

Then the next step is to find the displacement.

 

[aaronfue]

Don't forget the constant of integration.

Then the next step is to find the displacement.

s(t) = \frac{t^4}{3} - t² + c₁t + c₂

c₁ is the constant after integrating for v(t) & c₂ is the constant for s(t).

Do I have to set the constants equal to the given positions and t=0 and t=2?

 

[tms]

Do I have to set the constants equal to the given positions and t=0 and t=2?

No. You plug in the value for t, and then solve for the constants.

 

[aaronfue]

No. You plug in the value for t, and then solve for the constants.

Ok. I plugged in the t=0, s=1 and got c₂ = 1 (I think this would be -1 since the location is to the left of the origin?)

So I then plugged t=2, s= -20 and my c₁ = -9.84

Finally I plugged my c₁ and c₂ into my s(t) equation:

s(t) = \frac{1}{3}t⁴ - t² - 9.84t -1

And after using t=4.3, I got s = 52.16 m

Unfortunately, this is not correct. The correct answer was 50.8 m. I'm not sure where I messed up, but I'm thinking it was my rounding.

 

[aaronfue]

Ok. I plugged in the t=0, s=1 and got c₂ = 1 (I think this would be -1 since the location is to the left of the origin?)

So I then plugged t=2, s= -20 and my c₁ = -9.84

Finally I plugged my c₁ and c₂ into my s(t) equation:

s(t) = \frac{1}{3}t⁴ - t² - 9.84t -1

And after using t=4.3, I got s = 52.16 m

Unfortunately, this is not correct. The correct answer was 50.8 m. I'm not sure where I messed up, but I'm thinking it was my rounding.

Found my mistake! I was subtracting my numbers from +20 and not using -20 when calculating my c₁!

I got:
s = 50.75 = 50.8 m!

Dang!