Higher June 2025 Paper 6 Q12
12 You may use these kinematics formulae to answer this question.
\(v = u + at\)
\(s = ut + \frac{1}{2}at^2\)
where \(a\) is the constant acceleration, \(u\) is the initial velocity, \(v\) is the final velocity, \(s\) is the displacement and \(t\) is the time taken.
A particle has an initial velocity of 2 m/s.
After 6 seconds the particle has a velocity of 11 m/s.
The particle had constant acceleration.
Work out the distance the particle has travelled in the 6 seconds. [4]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| 39 | 4 | B2 for [\(a\) =] 1.5 oe or M1 for \(11 = 2 + 6a\) or better | |
| M1 for \(2 \times 6 + \frac{1}{2} \times\) their a \(\times 6^2\) or better | their a must come from use of \(v = u + at\) | ||