# Graph of distance travelled in $n$th second with time

I was asked to graph the distance travelled in $$n$$th second by an object with time, taking the object to be at rest initially, and accelerating with a constant acceleration of $$1 \, \text{m}/\text{s}^2$$.

According to the formula, $$s= u + \frac{1}{2}a(2t-1)$$ the y-intercept comes out to be $$\frac{-1}{2}$$.

I'm having trouble understanding how the distance travelled can be negative. Any help is appreciated!

Let's break down how this equation was derived.

This equation comes from the relation $$s=ut+\frac 12at^2$$. Let's define it as $$s(t)$$ and begin.

To work out the distance traveled at a particular second, we can think of taking the current distance $$s(t)$$ gives us and subtracting the distance from the previous second.

So our new formula, let's call it $$d(t)$$ can be derived by simplifying $$s(t)-s(t-1)$$ which equals $$at+u-\frac a2$$.

If we factor $$\frac 12a$$ we get $$u+\frac 12a(2t-1)$$ which is what you have.

Now knowing how this equation works, let's try reason how $$d(0)$$ could operate and why it makes no sense, hence unreasonable to use practically.

Mathematically, you can't have $$s(0)-s(-1)$$, it just doesn't make sense, you can't have negative time.

Logically, before the object has even started moving, is it fair to ask how much it has traveled in this second? We know how this equation works, so we can reasonably say we cannot rely on it to calculate the distance traveled at the 0th second.

• thank you so much! :-) Dec 23, 2019 at 11:15