Solving for time when delta x=0?

[atypical]

So my physics book says "from x(final)=x(initial)+V(initial)t+1/2at^2 observe that when x(final)=x(initial), the time is given by t=-2V(initial)/a"
Now if change in x is equal to 0, wouldn't your velocity v(initial)t from above along with your acceleration also be zero?
How is the book getting the t equation when delta x is zero?

 

[fzero]

What if x is the height of an object thrown into the air?

 

[atypical]

What if x is the height of an object thrown into the air?

ok, so what I said is correct for horizontal movement but not necessarily true for vertical moment, correct? I still do not see how the book is getting the time formula from observing that equation.

 

[D H]

Now if change in x is equal to 0, wouldn't your velocity v(initial)t from above along with your acceleration also be zero?

That is one solution, yes. It is the trivial solution.

How is the book getting the t equation when delta x is zero?

In general you have

$$\frac 1 2 at^2 + v_0t + (x_0-x_f) = 0$$

That is a quadratic equation in t. Quadratic equations have two solutions. In the special case x₀=x_f this reduces to

$$\frac 1 2 at^2 + v_0t = 0$$

Factoring out the common t on the left-hand side,

$$(t)(\frac 1 2 at + v_0) = 0$$

So the two solutions are given by $t=0$ and $1/2 a t +v_0 = 0$.

 

[atypical]

Ok, that made complete sense. I am a bit rusty on my math and forgot when you factor "t" out you get two solutions. But i am having trouble visually imagining this. Say you visually imagine x position graph. If there is no change in x horizontally (when x(initial)=x(final)), you would have a constant line between your initial and final values of x in that time interval. Now, if your position remains a constant, then you can't have any slope therefore your velocity and acceleration would have to be zero. What am i overlooking?

 

[fzero]

Ok, that made complete sense. I am a bit rusty on my math and forgot when you factor "t" out you get two solutions. But i am having trouble visually imagining this. Say you visually imagine x position graph. If there is no change in x horizontally (when x(initial)=x(final)), you would have a constant line between your initial and final values of x in that time interval. Now, if your position remains a constant, then you can't have any slope therefore your velocity and acceleration would have to be zero. What am i overlooking?

The position is not a constant, while

$$x(t=t_i) = x(t=t_f),$$

for times $$t_i<t<t_f$$ it's given by $$x(t)$$. What is physically happening is that the acceleration $$a$$ must act in the opposite direction from the initial velocity $$v_i$$. At some time $$t=t_*$$, the velocity $$v(t_*)=0$$ and after that the object is pushed back until it's at its initial position.

The formula doesn't care whether or not this is horizontal or vertical motion as long as the acceleration is constant. The case of vertical motion where $$a=-g$$ is due to gravity is just the clearest real-life example.