Solving kinematics problems step by step — Physics

Before this

This leads to

The same ball, the whole story

In the lecture on velocity-time graphs, a ball was thrown straight up at 15 m/s. From the graph, we found that it reached the top after about 1.53 s, about 11.5 m up. But we stopped there. How long is it in the air altogether? How fast is it moving when it comes back to the thrower's hand?

Questions like these feel messy. The ball goes up, stops, and comes down. Its velocity changes sign halfway through. Many students know all five SUVAT equations and still get these questions wrong. The reason is almost never the algebra. It is the signs.

This lecture gives you a method that removes that problem. It has three steps, and the first one is the one that most students skip.

Step 1: choose a positive direction, and write it down

Velocity, displacement and acceleration all have directions. A number like 15 m/s means nothing on its own when the motion can go both ways. So before you write any equation, decide which direction counts as positive. Draw an arrow and label it.

Suppose you choose up as positive. Then everything follows from that one choice:

A ball thrown upward has a positive starting velocity: $u = + 15$ m/s.
Gravity pulls down, which is the negative direction. So $a = - 9.8$ m/s², on the way up and on the way down.
A position above the starting point is a positive displacement. A position below it is negative.

Up is positive. The velocity is positive on the way up, zero for an instant at the top, and negative on the way down. The acceleration is −9.8 m/s² at every moment.

There is no "correct" direction. You may choose down as positive instead, and the physics will agree. What matters is that you choose once and never change in the middle of a problem. A sign that flips halfway through quietly ruins every answer after it.

About $g$ : this lecture uses $g = 9.8$ m/s². Some exam questions tell you to use 10 m/s². That is a rounding, used to make the arithmetic easier. Always use the value the question gives.

Steps 2 and 3: list all five letters, then choose the equation

Each SUVAT equation contains four of the five quantities and leaves one out. So the method is:

The method

Step 1. Choose the positive direction. Write it down.

Step 2. Write all five letters, $s$ , $u$ , $v$ , $a$ and $t$ , and fill in what you know, with signs. Mark the one you want with "?" and the one you neither know nor want with "—".

Step 3. Choose the equation that leaves out the "—" letter. Solve. Then check that the answer makes physical sense.

Here are the five equations again, each labelled with the letter it leaves out:

Equation	Leaves out
$v = u + a t$	$s$
$s = \frac{1}{2} (u + v) t$	$a$
$s = u t + \frac{1}{2} a t^{2}$	$v$
$s = v t - \frac{1}{2} a t^{2}$	$u$
$v^{2} = u^{2} + 2 a s$	$t$

Step 2 looks slow. It is not. It takes twenty seconds, and it makes the choice of equation automatic. It also forces you to think about every sign before you calculate.

Before the worked examples, test your sense of signs on the ball.

Predict, then reveal — back at the hand

The ball is thrown up at 15 m/s and caught at the same height. Up is positive. What is its velocity just as it reaches the hand again?

The question: a ball is thrown up at 15 m/s and caught at the same height, with up as positive. What is its velocity when it returns?

Answer: −15 m/s. From $v^{2} = u^{2} + 2 a s$ with $s = 0$ : $v^{2} = 1 5^{2}$ , so $v = \pm 15$ . The ball is moving down, so $v = - 15$ m/s. It returns at the same speed it was thrown with, in the opposite direction.

Notice that $v^{2} = u^{2} + 2 a s$ gave two answers, +15 and −15. The equation only knows about squares, so it cannot tell up from down. You must choose the sign that matches the motion. This happens often with vertical motion, so always ask: which way is it moving at this moment?

Worked examples

Worked Example 1 — The ball thrown up

A ball is thrown straight up at 15 m/s from a hand. Find (a) its maximum height and (b) the time until it returns to the hand. Take $g = 9.8$ m/s².

Step 1. Up is positive.

(a) Step 2. $s = ?$ , $u = + 15$ , $v = 0$ (at the top it stops for an instant), $a = - 9.8$ , $t =$ —.

Step 3. Leave out $t$ : use $v^{2} = u^{2} + 2 a s$ .

0 = 1 5^{2} + 2 (- 9.8) s \Rightarrow 19.6 s = 225 \Rightarrow s \approx 11.5 m

(b) Step 2. Back at the hand, the displacement is zero. $s = 0$ , $u = + 15$ , $a = - 9.8$ , $t = ?$ , $v =$ —.

Step 3. Leave out $v$ : use $s = u t + \frac{1}{2} a t^{2}$ .

0 = 15 t - 4.9 t^{2} = t (15 - 4.9 t)

So $t = 0$ or $t = 15 \div 4.9 \approx 3.06$ s. The answer $t = 0$ is the moment of throwing, so the ball returns after about 3.06 s.

Check: the time to the top is $15 \div 9.8 \approx 1.53$ s, and $2 \times 1.53 = 3.06$ s. Without air resistance, the way down takes exactly as long as the way up.

Worked Example 2 — A stone dropped from a cliff

A stone is dropped from the top of a cliff 45 m high. Find the time it takes to hit the ground and its speed on impact.

Step 1. This time, choose down as positive. Everything moves down, so all the numbers become positive.

Step 2. $s = + 45$ , $u = 0$ ("dropped" means it starts from rest), $a = + 9.8$ , $t = ?$ , $v = ?$ .

Step 3. For $t$ , leave out $v$ : $45 = 0 + 4.9 t^{2}$ , so $t^{2} \approx 9.18$ and $t \approx 3.03$ s.

For $v$ , leave out $t$ , so that an early rounding cannot spread: $v^{2} = 0 + 2 (9.8) (45) = 882$ , so $v \approx 29.7$ m/s.

Check with the other route: $v = u + a t = 9.8 \times 3.03 \approx 29.7$ m/s. The two routes agree.

Worked Example 3 — Thrown up from a balcony

A ball is thrown straight up at 12 m/s from a balcony 20 m above the ground. When does it hit the ground?

Step 1. Up is positive, with $s = 0$ at the balcony.

Step 2. The ground is 20 m below the start, so $s = - 20$ . $u = + 12$ , $a = - 9.8$ , $t = ?$ , $v$ unknown.

Using $s = u t + \frac{1}{2} a t^{2}$ would give a quadratic equation. There is a simpler route: find $v$ first, then $t$ .

v^{2} = 1 2^{2} + 2 (- 9.8) (- 20) = 144 + 392 = 536 \Rightarrow v \approx \pm 23.15

The ball is moving down when it lands, so $v \approx - 23.15$ m/s. Now use $v = u + a t$ :

t = \frac{v - u}{a} = \frac{- 23.15 - 12}{- 9.8} \approx 3.59 s

Check: the ball first rises for $12 \div 9.8 \approx 1.22$ s. So it spends about 2.37 s coming down from its highest point, which is longer than the 1.22 s it spent going up. That makes sense, because it falls further than it rose: past the balcony and on to the ground.

Worked Example 4 — Two stages: thinking, then braking

A driver travelling at 20 m/s sees a hazard. It takes her 0.8 s to react, and then the brakes slow the car at 8 m/s². Find the total stopping distance.

The motion has two stages with different accelerations. Each stage needs its own list.

Stage 1 (reacting). No braking yet, so the speed stays 20 m/s: $s = 20 \times 0.8 = 16$ m.

Stage 2 (braking). Forward is positive. $u = + 20$ , $v = 0$ , $a = - 8$ , $s = ?$ , $t =$ —. Use $v^{2} = u^{2} + 2 a s$ : $0 = 400 - 16 s$ , so $s = 25$ m.

Total stopping distance $= 16 + 25 = 41$ m.

Check: the final velocity of stage 1 (20 m/s) is the starting velocity of stage 2. That link is how multi-stage problems join together.

Now watch the signs change in real time. Choose a launch speed and run the ball.

Simulation — a ball thrown straight up

Choose the launch speed. Press Play, or Step through 0.25 s at a time. Watch the signs of s and v. Up is positive.

Launch speed u: 15 m/s

The ball thrown up at 15 m/s, every half second, with up as positive ( $s = 15 t - 4.9 t^{2}$ , $v = 15 - 9.8 t$ ):

t (s)	s (m)	v (m/s)	What is happening
0	0	+15	just thrown, moving up
0.5	6.28	+10.1	rising, slowing down
1.0	10.1	+5.2	rising, slowing down
1.53	≈ 11.5	≈ 0	at the top, stopped for an instant
2.0	10.4	−4.6	falling, speeding up
2.5	6.88	−9.5	falling, speeding up
3.06	≈ 0	≈ −15	back at the hand

The acceleration is −9.8 m/s² in every row.

Look at what the simulation shows. On the way up, $v$ is positive and $a$ is negative: they point opposite ways, so the ball slows down. On the way down, both are negative: they point the same way, so the ball speeds up. At the top, $v$ is zero but $a$ is still −9.8 m/s². That is it. That is the whole idea of sign convention: one choice of direction, and every sign then tells the truth.

Stop and think

A student solves Worked Example 1(b) with $s = 11.5$ m instead of $s = 0$ , "because the ball travels 11.5 m". What has gone wrong?

Show the answer

The student has confused displacement with distance. $s$ in SUVAT is the displacement: where the ball is compared with where it started. When it is back in the hand, it is exactly where it started, so $s = 0$ . The distance travelled is 23 m (11.5 m up and 11.5 m down), but distance never appears in the SUVAT equations. Using $s = 11.5$ would find the time to reach the top, which is a different question.

Three classic mistakes

Trap 1 — Writing a = +9.8 when up is positive

If up is positive, gravity is negative: $a = - 9.8$ m/s², on the way up and on the way down. The most common sign error in mechanics is writing $+ 9.8$ out of habit. Look at your arrow before you write $a$ .

Trap 2 — Taking the wrong root

$v^{2} = u^{2} + 2 a s$ always gives two possible values, $+ v$ and $- v$ . The equation cannot tell you the direction. Choose the sign from the physical motion: a ball landing is moving down.

Trap 3 — Mixing units

Speeds are often given in km/h. SUVAT needs metres and seconds. Divide km/h by 3.6 to get m/s before you start: 72 km/h is 20 m/s.

Try it yourself

Work each one on paper first. Use $g = 9.8$ m/s². Open a hint only when you are stuck.

A stone is thrown straight down at 5 m/s from a bridge 30 m above the water. Find its speed as it hits the water and the time it takes.

Hint 1
Everything moves down, so choose down as positive. Then $u = + 5$ , $a = + 9.8$ , $s = + 30$ .

Hint 2
Find $v$ first with $v^{2} = u^{2} + 2 a s$ , then $t$ from $v = u + a t$ .

Full answer
$v^{2} = 25 + 2 (9.8) (30) = 613$ , so $v \approx 24.8$ m/s. Then $t = (24.8 - 5) \div 9.8 \approx 2.02$ s. Check: $s = u t + \frac{1}{2} a t^{2} = 5 (2.02) + 4.9 (2.02)^{2} \approx 10.1 + 20.0 = 30.1$ m, which is 30 m within rounding.
A ball is thrown straight up at 20 m/s. Find its maximum height, and its displacement and velocity after 3 s.

Hint 1
Up is positive, so $a = - 9.8$ . At the top, $v = 0$ .

Hint 2
For 3 s, use $s = u t + \frac{1}{2} a t^{2}$ and $v = u + a t$ . Expect $v$ to be negative: is the ball past the top?

Full answer
Height: $0 = 400 - 19.6 s$ , so $s \approx 20.4$ m. After 3 s: $s = 20 (3) - 4.9 (9) = 60 - 44.1 = 15.9$ m, and $v = 20 - 9.8 (3) = - 9.4$ m/s. The ball is 15.9 m up and falling. Check: the top is at $20 \div 9.8 \approx 2.04$ s, so at 3 s it has passed the top, which matches the negative $v$ .
A train travelling at 30 m/s brakes steadily and stops in 450 m. Find its deceleration and the time it takes to stop.

Hint 1
Forward is positive. $u = + 30$ , $v = 0$ , $s = + 450$ . Which letter is not needed for the deceleration?

Hint 2
Use $v^{2} = u^{2} + 2 a s$ for $a$ , then $v = u + a t$ for $t$ .

Full answer
$0 = 900 + 2 a (450)$ , so $a = - 1$ m/s²: a deceleration of 1 m/s². Then $0 = 30 - 1 \times t$ , so $t = 30$ s. Check with $s = \frac{1}{2} (u + v) t = \frac{1}{2} (30) (30) = 450$ m. Correct.
A model rocket rises straight up from rest with an acceleration of 4 m/s² for 10 s. Then its fuel runs out, and only gravity acts. Find its greatest height.

Hint 1
This is two stages. Stage 1 ends when the fuel runs out. Find the height and velocity at that moment.

Hint 2
Stage 1: $v = 0 + 4 \times 10 = 40$ m/s and $s = \frac{1}{2} \times 4 \times 1 0^{2} = 200$ m. Stage 2 starts at 40 m/s with $a = - 9.8$ , and ends at the top with $v = 0$ .

Full answer
Stage 1: $v = 40$ m/s and $s = 200$ m. Stage 2: $0 = 4 0^{2} + 2 (- 9.8) s$ , so $s = 1600 \div 19.6 \approx 81.6$ m. Greatest height $\approx 200 + 81.6 = 281.6$ m. Notice that the rocket keeps rising after the fuel runs out, for $40 \div 9.8 \approx 4.1$ s more. Check: stage 2's starting velocity, 40 m/s, is stage 1's final velocity.

The bigger idea

The method in this lecture is not really about SUVAT. It is about turning a story into a list of facts before calculating anything. Choose the frame, write down what you know with its sign, and only then pick a tool.

You will use exactly this habit in projectile motion, the next topic. There, an object moves sideways and up-and-down at the same time, and the trick is to treat the two directions as two separate SUVAT lists. With a clear frame, a problem that looks impossible becomes two problems you can already solve.

Key concepts in this lesson

Before using any equation, choose a positive direction and write it down, then keep it for the whole problem.
With up as positive, the acceleration due to gravity is −9.8 m/s² at every moment, on the way up, at the top and on the way down.
Listing all five SUVAT letters with signs shows which letter is not needed, and the right equation is the one that leaves it out.
The equation $v^{2} = u^{2} + 2 a s$ gives two signs for $v$ , so the correct sign must be chosen from the direction of motion.
A problem with stages is solved one stage at a time, with the final velocity of one stage becoming the starting velocity of the next.