# Thread: Integration of Zero where d^2z/dt^2 = 0

1. ## Integration of Zero where d^2z/dt^2 = 0

Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
velocity ˆ−j +2ˆk and moves under a constant force F = −i +2ˆj. Find
the position and velocity for all time.

The Equation of motion ma = F implies that,

a = −i +2ˆj

which gives the three scalar equations:

d^2x/dt^2= −1,

d^2y/dt^2= 2,

d^2z/dt^2= 0

integrating for x yields:

x = −1/2t^2+1

integrating for y yields:

y = t^2− t

I am unable find z, where:

d^2z/dt^2 = 0

I need to integrate this equation twice to obtain:

z = 2t

But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.

Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.

2. Originally Posted by GrahamJamesK
Suppose a particle of mass 1 kg is initially at (1, 0, 0) and with initial
velocity ˆ−j +2ˆk and moves under a constant force F = −i +2ˆj. Find
the position and velocity for all time.

The Equation of motion ma = F implies that,

a = −i +2ˆj

which gives the three scalar equations:

d^2x/dt^2= −1,

d^2y/dt^2= 2,

d^2z/dt^2= 0

integrating for x yields:

x = −1/2t^2+1

integrating for y yields:

y = t^2− t

I am unable find z, where:

d^2z/dt^2 = 0

I need to integrate this equation twice to obtain:

z = 2t

But as far as i understand integrating 0 yields 0. I have a feeling that I'm overlooking something simple but at present it is escaping me and i cannot figure out how to arrive at the given value.

Any help or advice would be greatly appreciated as I am finding this area of Applied math confusing.
$\displaystyle \displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}$

If c = 2 and n = 0, you have your z = 2t

3. Originally Posted by dwsmith
$\displaystyle \displaystyle\int 0 \ dt=c\Rightarrow\int c \ dt=ct+n \ \ \ c,n\in\mathbb{R}$

If c = 2 and n = 0, you have your z = 2t
Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.

4. Originally Posted by GrahamJamesK
Thank you for your reply. I understand this, but how can I arrive at 2 if I assume that "z=2t" is unknown.
Why are you assuming it is 2t? If you know that, you just set them equal.

$\displaystyle ct+n=2t\Rightarrow c=2 \ \ n=0\Rightarrow z=2t$

5. Ok that makes sense, thank you.

6. From $\displaystyle \frac{d^2z}{dt^2}= 0$, you get $\displaystyle \frac{dz}{dt}= C$, a constant. But you are told that the velocity at t= 0 is -j+ 2k so that the z component of velocity is 2: the constant is 2. From $\displaystyle \frac{dz}{dt}= 2$, integrating again gives $\displaystyle z(t)= 2t+ D$. Since you wer told that the initial position was (1, 0, 0), z(0)= 2(0)+ D= 0 and so D= 0. z= 2t.

7. Your explanation was very instructive, I was getting confused by the vector labels and the axis labels. Kudos!