KCSE 2025 Maths P2 Q15 — Kinematics by Integration (Distance from Velocity)

KCSE 2025 Form 4 Calculus

Published 12 July 2026

The Question

“A particle starts from a point O and moves in a straight line. Its velocity V m/s at time t seconds is given by V = 4 - t. The distance S of the particle from O at time t = 2 seconds is 7 m. Calculate S when t = 4 seconds.”

Integrate the velocity

Velocity tells you how fast the distance is changing, so to get the distance back you integrate the velocity with respect to time. Integrate each term of 4 minus t, and remember to add a constant of integration c because integrating always leaves an unknown constant.

S = \int (4 - t)\, dt = 4t - \frac{t^{2}}{2} + c

Use the known point to find c

You are told the distance is 7 metres when the time is 2 seconds. Substitute t = 2 and S = 7 into the expression so the only unknown left is c, then solve for it.

7 = 4(2) - \frac{2^{2}}{2} + c

7 = 8 - 2 + c

7 = 6 + c \Rightarrow c = 1

Write the complete distance rule

Put the value of c back into the expression to get the full formula for the distance from O at any time t.

S = 4t - \frac{t^{2}}{2} + 1

Evaluate the distance at t = 4

Finally substitute t = 4 into the complete rule and simplify to find the distance from O at that instant.

S = 4(4) - \frac{4^{2}}{2} + 1

S = 16 - 8 + 1 = 9

Final Result

The distance S is 9 m when t = 4 seconds.

Why this method works

Displacement and velocity are linked by calculus: velocity is the rate at which distance changes, so integrating velocity reverses that and rebuilds the distance. Integration alone cannot pin down the starting offset, which is why a constant c appears; the given fact that S = 7 at t = 2 supplies exactly the one piece of information needed to fix c. Once the constant is known the rule is complete, and any time can be substituted to read off the distance.

✓

Test the rule at t = 2: S = 4(2) - 4/2 + 1 = 8 - 2 + 1 = 7 m, which matches the given information, so the formula is correct.