| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Find acceleration from distances/times |
| Difficulty | Easy -1.2 This is a straightforward mechanics question requiring only basic calculus (differentiation to find acceleration, integration to find distance) and SUVAT application. All steps are routine: reading initial velocity from the equation, differentiating a linear function, integrating a quadratic, and applying a standard SUVAT equation. No problem-solving insight or challenging manipulation is needed. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
1 A particle is travelling in a straight line. Its velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t$ seconds is given by
$$v = 6 + 4 t \quad \text { for } 0 \leqslant t \leqslant 5$$
(i) Write down the initial velocity of the particle and find the acceleration for $0 \leqslant t \leqslant 5$.\\
(ii) Write down the velocity of the particle when $t = 5$. Find the distance travelled in the first 5 seconds.
For $5 \leqslant t \leqslant 15$, the acceleration of the particle is $3 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(iii) Find the total distance travelled by the particle during the 15 seconds.
\hfill \mbox{\textit{OCR MEI M1 2009 Q1 [8]}}