The acceleration is about -2.22 m/s² toward the east, so the correct multiple-choice answer is -8 only if we are discussing a misprint or a different problem; for the stated values, none of the listed choices fits exactly. Here is why. The car starts at 40 km/h west and stops in 5 s. Converting 40 km/h to about 11.11 m/s and dividing the change of -11.11 m/s by 5 s gives -2.22 m/s². The negative sign means acceleration points opposite the initial westward motion.
The multiple-choice answer -8 would fit if the initial speed were 40 m/s, not 40 km/h, with the same 5 s stop. That distinction matters. Students often lose points because they skip the unit conversion. This explainer works through the mechanism, the numbers, the edge cases, and the exam traps behind this question.
The mechanism: why stopping means negative acceleration
Acceleration is the rate at which velocity changes with time. In symbols, average acceleration equals the change in velocity divided by the time interval. Velocity is a vector, so direction is part of the value. When the car moves west and slows down, its westward velocity shrinks toward zero.
If we call west the positive direction, the initial velocity is +11.11 m/s. The final velocity is 0 m/s. The change is 0 minus 11.11, or -11.11 m/s. Dividing by 5 s yields -2.22 m/s². The minus sign tells you the acceleration points east, opposite the westward motion.
That is the core mechanism. Acceleration does not always mean speeding up. It means any change in velocity, including slowing down or turning. A slowing car has acceleration directed opposite its motion. Physicists call this deceleration, but the formal name is still negative acceleration relative to the chosen positive direction.
One plain-language sentence: the car's speed drops by about 2.22 meters per second every second while it brakes.
The arithmetic: converting km/h to m/s cleanly
The single most common error is mixing units. The question asks for acceleration in m/s² but gives the speed in km/h. You must convert before dividing. One kilometer equals 1,000 meters, and one hour equals 3,600 seconds. So 40 km/h equals 40 times 1,000 divided by 3,600, which is 11.11 m/s.
Here is the full chain. Start speed is 40 km/h west, or 11.11 m/s west. End speed is 0 m/s. Time is 5 s. Acceleration is (0 - 11.11) / 5, which equals -2.22 m/s². If west is positive, the answer is -2.22 m/s². If east is positive, the sign flips to +2.22 m/s², but the physical meaning is identical: the acceleration points east.
Compare that to the answer -8. If a student forgets to convert and treats 40 as meters per second, the calculation becomes (0 - 40) / 5, which equals -8 m/s². That is exactly how the distractor answer arises. The -8 choice is a trap for students who skip the conversion step.
A quick sanity check helps. A car slowing from about 11 m/s to zero in 5 s loses roughly 2.2 m/s each second. That is a gentle, realistic braking rate. Losing 8 m/s each second would be a hard, emergency-style stop. The physical feel confirms which answer is reasonable.
A timeline of the stop
The table below traces the car's velocity second by second, assuming constant acceleration. This is a model, not a measurement, but it shows how the numbers behave.
| Time (s) | Velocity (m/s, west) | Change from prior second (m/s) |
|---|---|---|
| 0 | 11.11 | — |
| 1 | 8.89 | -2.22 |
| 2 | 6.67 | -2.22 |
| 3 | 4.44 | -2.22 |
| 4 | 2.22 | -2.22 |
| 5 | 0.00 | -2.22 |
Every second the westward velocity falls by the same 2.22 m/s. That constant drop is what "constant acceleration" means. The car covers distance during the stop, but the question only asks about acceleration, so distance is not needed here.
If you wanted stopping distance as a cross-check, the average velocity during a constant-acceleration stop is half the starting velocity, about 5.56 m/s. Multiply by 5 s and you get about 27.8 m. That is roughly the length of two and a half school buses. It is a useful reality check on whether the braking rate is plausible.
Edge cases: when the simple model breaks
The formula assumes constant acceleration. Real braking is rarely perfectly constant. A driver eases onto the pedal, the brakes warm up, and the tires interact with changing road surfaces. If braking force varies, the average acceleration over 5 s can still be -2.22 m/s², but the instantaneous value at any moment may differ.
Direction conventions create another edge case. The answer -2.22 m/s² is correct only if west is positive. Many textbooks define the direction of motion as positive by default, which produces a negative result for a stop. But some problems set east as positive, and then the same stop yields +2.22 m/s². Always state your sign convention.
Significant figures matter too. The inputs are 40 km/h and 5 s. The 40 has one or two significant figures, and the 5 has one. A strict treatment rounds the answer to about -2 m/s². Most multiple-choice exams expect -2.22 m/s², so follow your instructor's convention.
There is also the question of whether the car stops exactly at 5 s or begins stopping at 5 s. The wording "comes to a stop in 5 s" means the stop is complete at the 5-second mark. Read carefully. A different phrasing could change the time interval and therefore the answer.
A comparison of the answer choices
The table below shows what each choice implies and why it is right or wrong for this problem.
| Choice | What it implies | Verdict for 40 km/h in 5 s |
|---|---|---|
| 8 | Forgot the negative sign; treated 40 as m/s | Wrong sign and wrong units |
| 200 | Multiplied instead of divided, or misread units | Wrong operation |
| 0.125 | Divided time by speed or inverted the ratio | Wrong operation |
| -8 | Treated 40 as m/s and kept the negative sign | Right method, wrong unit conversion |
| -2.22 (not listed) | Correct conversion and correct sign | Correct physics answer |
The -8 choice is the most instructive. It shows a student who understands the sign convention and the division but skipped the km/h to m/s conversion. That is a near-miss, not a random error. If the problem had said 40 m/s, -8 would be exactly right.
Misconceptions and red flags
Several beliefs trip students up on this exact question. Watch for these warning signs in your own work.
Treating km/h and m/s as interchangeable. They are not. A factor of 3.6 separates them. Assuming a negative acceleration always means slowing down. It means acceleration points opposite the chosen positive direction. Forgetting that a stop is a change in velocity, not just a change in speed. Direction is part of velocity. Dropping the sign because "acceleration can't be negative." It can, and the sign carries real meaning. Rounding 40 km/h to 10 m/s and expecting an exact match. The precise value is 11.11 m/s, and the difference changes the answer.
The counterintuitive detail: the answer -8 is not nonsense. It is the correct answer to a slightly different problem. That is why it appears as a distractor. Exam writers build wrong choices from common near-mistakes, and this one is engineered to catch a specific error.
A professional caveat: in real vehicle dynamics, braking acceleration is often quoted in g, where 1 g is about 9.81 m/s². A deceleration of 2.22 m/s² is about 0.23 g. Normal passenger-car braking on dry pavement can reach roughly 0.7 to 0.9 g, so this stop is gentle. That context helps you judge whether a computed number is physically sensible.
Practical actions for solving this question
Use a repeatable procedure so unit conversions and signs do not slip.
Write down what is given with units. Here: 40 km/h west, final speed 0, time 5 s. Convert every speed to m/s before doing anything else. Divide km/h by 3.6. Choose a positive direction and write it down. West positive is the natural choice here. Compute the change in velocity. Final minus initial, keeping signs. Divide by the time interval. Average acceleration equals change in velocity over time. Check plausibility. Does the braking rate feel gentle, moderate, or extreme?
Apply the procedure: 40 km/h becomes 11.11 m/s. West positive, so initial is +11.11 and final is 0. Change is -11.11. Divide by 5 s to get -2.22 m/s². The rate is gentle, which matches ordinary braking.
Before an exam, practice the conversion until it is automatic. Common anchors help: 36 km/h equals 10 m/s, 72 km/h equals 20 m/s, and 100 km/h equals about 27.8 m/s. Knowing these by heart lets you spot unit errors quickly. The 40 km/h value sits a little above 36 km/h, so the answer should be a little above 10 m/s, near 11.1 m/s.
Limitations: what this problem does not tell you
The calculation gives average acceleration over 5 s under an assumed constant rate. It says nothing about peak braking force, tire grip, road grade, or driver reaction time. Those factors shape real stopping events but are outside the scope of this question.
The problem also ignores air resistance and rolling resistance. At low speeds these are minor, but at highway speeds they matter. A full vehicle-dynamics model would separate brake force from drag and account for weight transfer between axles. None of that changes the exam answer, but it matters in engineering.
Finally, the answer depends on the sign convention you adopt. There is no single "true" sign without a stated positive direction. The magnitude, about 2.22 m/s², is fixed. The sign is a bookkeeping choice. State it explicitly so your answer is unambiguous.
For this multiple-choice question, the listed choices do not include the correct value. The answer -8 corresponds to a 40 m/s initial speed, not 40 km/h. If your exam forces a choice among these options, flag the unit mismatch and show your conversion. The physics answer is about -2.22 m/s² west-positive, which rounds to about -2 m/s² at the given significant figures.