Let's denote the width of the rectangle as $w$ and the length as $l$. According to the problem, the length exceeds the width by 5 units. Therefore, we can write:

$l = w + 5$

The formula for the perimeter $P$ of a rectangle is:

$P = 2l + 2w$

Substitute $l = w + 5$ into the perimeter formula:

$P = 2(w + 5) + 2w$

Now, simplify:

$P = 2w + 10 + 2w = 4w + 10$

So, the perimeter of the rectangle is:

$P = 4w + 10$

Do you need further details or explanations on this?

Here are 5 related questions for further practice:

1. If the width $w$ is 7 units, what is the perimeter of the rectangle?
2. If the perimeter is 50 units, what is the width of the rectangle?
3. How does the formula for perimeter change if the length exceeds the width by 7 units instead of 5?
4. If the width $w$ is doubled, how does that affect the perimeter?
5. How would you calculate the area of this rectangle?

Tip: When working with formulas, always substitute known values carefully to avoid errors.