Solving for time in the distance rate formula

Key concept: This problem asks you to rearrange a linear formula so that time is written alone on one side.

Step by step

Identify the goal

The equation is given as $s = d + vt$, and the task is to solve for $t$. That means we must isolate the variable $t$ using inverse operations. In algebra, the goal is always to undo whatever is attached to the target variable while keeping the equation balanced.

Here, $d$ is added to $vt$, so the first step is to subtract $d$ from both sides:

$s - d = vt$

Now $t$ is still multiplied by $v$, so divide both sides by $v$:

$t = \frac{s-d}{v}$

Step-by-step method

Start with the original equation:

$s = d + vt$

Subtract $d$ from both sides:

$s - d = vt$

Then divide both sides by $v$:

$\frac{s-d}{v} = t$

Rewrite in standard order:

$t = \frac{s-d}{v}$

This matches choice (b) in the list, even though the repeated line at the bottom shows the algebraic result in another form. The key idea is that you must remove addition before removing multiplication.

Why the other choices are wrong

Choice $(a)$, $t = \frac{s}{d}$, divides by the wrong quantity. It treats $d$ as if it were the coefficient of $t$, but it is not.

Choice $(c)$, $t = \frac{s}{v} - d$, changes the order of operations incorrectly. You cannot divide only part of the expression and then subtract $d$ afterward unless the original equation has that structure.

Choice $(d)$, $t = s - v$, ignores both the added term $d$ and the multiplication by $v$.

Common algebra property to remember

When a variable is inside a product, divide to undo multiplication. When a term is added or subtracted, use the opposite operation to remove it. This same strategy works for many formulas, not just this one.

A good habit is to check your final answer by substituting it back into the original equation. If you do that here, the left and right sides stay equal, confirming that the rearrangement is correct.

Pitfall alert

A common mistake is to divide by the wrong term because the equation contains several symbols. In $s = d + vt$, only $v$ is multiplying $t$; $d$ is a separate added term. Another frequent error is forgetting to subtract $d$ first. If you try to divide immediately, you will not isolate $t$ correctly. Always reverse operations in the opposite order from how they are applied.

Try different conditions

If the formula were written as $s = vt + d$, the solving process would be exactly the same because addition is commutative. You would still subtract $d$ first and then divide by $v$, giving $t = \frac{s-d}{v}$. If the equation instead were $s = d - vt$, then the sign would change during the isolating step, and the result would be $t = \frac{d-s}{v}$. The structure of the formula determines the algebraic sign.

Further reading

isolating the variable, inverse operations, linear formula rearrangement