Apply first equation of motion. Find final velocity given u, a, t.

Find Velocity Using v = u + at — Kinematics Equation 1

This is the entry-level numerical for kinematics — but it's also where students learn the correct approach that will carry them through JEE numericals later. Write down knowns, identify the right equation, substitute carefully, and state the answer with units. Don't skip any step even when the problem looks trivial.

Question

A car starts from rest and accelerates uniformly at 2 m/s² for 10 seconds. Find the final velocity of the car.

Solution — Step by Step

Step 1: Write down the given information.

Initial velocity: $u = 0$ m/s (starts from rest)
Acceleration: $a = 2$ m/s²
Time: $t = 10$ s
Find: Final velocity $v = ?$

Step 2: Choose the correct equation.

We know $u$ , $a$ , and $t$ . We need $v$ . The equation connecting these four quantities is the first equation of motion:

$v = u + at$

Step 3: Substitute and calculate.

$v = 0 + 2 \times 10$

$v = 20 \text{ m/s}$

Answer: The final velocity of the car is 20 m/s.

Why This Works

The first equation of motion comes directly from the definition of acceleration:

$a = \frac{v - u}{t}$

Rearranging: $v = u + at$

That's all it is — acceleration times time gives the change in velocity, and we add that to the initial velocity to get the final velocity. The car started at 0, gained 2 m/s every second for 10 seconds, so it ends up at 20 m/s. Clean and logical.

📌 Note

"Starts from rest" always means $u = 0$ . This phrase appears in hundreds of kinematics problems. Recognize it immediately and write $u = 0$ without hesitation.

Bonus: Find Distance Covered Too

Since we have $u$ , $a$ , $t$ , and now $v$ , let's also find the distance for extra practice:

Using second equation of motion:

$s = ut + \frac{1}{2}at^2 = 0 + \frac{1}{2} \times 2 \times 10^2 = 100 \text{ m}$

Or using the third equation:

$v^2 = u^2 + 2as$

$400 = 0 + 2 \times 2 \times s$

$s = \frac{400}{4} = 100 \text{ m}$

Both give 100 m. Always good to verify using two methods when practicing.

Alternative Method — Direct Proportionality

Since $u = 0$ , the equation $v = at$ shows that velocity is directly proportional to time. We can use this conceptually:

At $t = 1$ s: $v = 2$ m/s

At $t = 5$ s: $v = 10$ m/s

At $t = 10$ s: $v = 20$ m/s

The velocity increases linearly from 0 to 20 m/s over 10 seconds. On a v-t graph, this is a straight line through the origin with slope $= 2$ m/s².

💡 Expert Tip

Always sketch the v-t graph mentally even for simple problems. It reinforces what the equations mean physically. For this problem: straight line from $(0, 0)$ to $(10, 20)$ . Slope $=$ acceleration $= 2$ m/s². Area under the line $=$ distance $= \dfrac{1}{2} \times 10 \times 20 = 100$ m.

Common Mistake

⚠️ Common Mistake

The most common error at this level is forgetting to write $u = 0$ when the problem says "starts from rest." Students sometimes write $v = at$ correctly but then plug in $u$ as some non-zero value from a previous part of the question. Each part of a problem has its own initial conditions. "Starts from rest" in this part means $u = 0$ for this part specifically.

A second mistake is confusing $a = 2$ m/s² (acceleration) with $u = 2$ m/s (initial velocity). Keep your variables clearly labelled. In exams, a small labelling error at the start leads to a completely wrong answer at the end.