KCSE 2025 Physics P1 Q11 — Uniform Deceleration (v² = u² + 2as)

KCSE 2025 Form 4 Kinematics

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The Question

“An object moving with an initial velocity of 3 m/s decelerates uniformly and comes to rest after travelling a distance of 10 m. Determine its acceleration.”

1

List what you are given

Write down the three quantities the question hands you. The object starts at 3 m/s, so the initial velocity u is 3 m/s. It comes to rest, so the final velocity v is 0. The distance travelled while stopping, s, is 10 m. The acceleration a is the unknown we want.

u=3 m/s,v=0,s=10 mu = 3\ \text{m/s},\quad v = 0,\quad s = 10\ \text{m}
2

Choose the right equation of motion

You are given u, v and s, and you want a — but you are not told the time. The equation of motion that links exactly those four quantities without time is v² = u² + 2as, so it is the natural choice here.

v2=u2+2asv^2 = u^2 + 2as
3

Substitute the values

Put the numbers straight into the equation, keeping every quantity in SI units so the answer comes out in m/s².

02=32+2×a×100^2 = 3^2 + 2 \times a \times 10
4

Simplify the arithmetic

Square the numbers and multiply out the constant term. Three squared is 9, and 2 × 10 is 20, which leaves a simple linear equation in a.

0=9+20a0 = 9 + 20a
5

Make a the subject and solve

Move the 9 across to the other side and divide by 20. The result is negative, which is exactly what we expect for something that is slowing down.

a=−920a = -\frac{9}{20}
a=−0.45 m/s2a = -0.45\ \mathrm{m/s^2}

Final Result

The acceleration is −0.45 m/s². The negative sign shows it is a deceleration — the object is slowing down.

Why this method works

The three equations of motion each connect a different set of the five kinematic quantities (u, v, a, s, t). Here the clock time is never mentioned, so any equation containing t would introduce a second unknown and leave us stuck. The equation v² = u² + 2as is the only one that ties together initial velocity, final velocity, acceleration and distance while leaving time out entirely — which is precisely the information we have. The minus sign in the answer is not a mistake to be dropped: acceleration is a vector, and a negative value simply means it points opposite to the motion, i.e. the object is decelerating.

✓

Check by working forwards: with a = −0.45 m/s², v² = 3² + 2(−0.45)(10) = 9 − 9 = 0, so the object does indeed come to rest after 10 m.