| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question requiring integration of velocity to find displacement, identifying when velocity changes sign, and relating acceleration to time. While it involves multiple parts and careful attention to distance vs displacement, the techniques are routine applications of calculus to kinematics with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| i \(x = [t^2 - 9 dt\) \(x = t^3/3 - 9t (+c)\) Finds x(2) Displacement = 15⅔ m OR \(x(2) = [t^3/3 - 9t]_0^2\) Displacement = 15⅔ m | M1* A1 D*M1 B1 [4] | Uses integration of v(t) Award if +c omitted Allow +c or c omitted Accept 15.3, 46/3. Must be +ve |
| ii \(t=0\) \(x=0\) or \(s=46/3\) hence x(0) or c = 0 or 46/3 Solves \(t^2 – 9 = 0\) \(t = (\pm)3\) \(x(3) = 3'/3 -9x3 (+ 15.3)\) \(x(3) = -18\) (or -2.67) Dist = 18 m | B1* M1 A1 D*M1 M1 D*B1 [6] | Needs explanation, may be seen in part i May be implied Value of t when direction of motion changes Substitutes cv(t) >2 in integrated x(t) Evaluates c – 18 may be implied award if … Accept 18(,0) [c=0 assumed] |
| iii \(a = d(t^2 - 9)/dt\) \(a = 2t\) 10 = 2t \(t = 5\) \(x(5) = (= 5^3/3 -9x5 + 15.3) = 12 \text{ m}\) OR \([t^3/3 - 9t]_3^5 = 12 \text{ m}\) | M1* A1 D*M1 A1 A1 [5] | Uses differentiation of v(t) |
| **Answer/Working** | **Marks** | **Guidance** |
|---|---|---|
| **i** $x = [t^2 - 9 dt$ $x = t^3/3 - 9t (+c)$ Finds x(2) Displacement = 15⅔ m OR $x(2) = [t^3/3 - 9t]_0^2$ Displacement = 15⅔ m | M1* A1 D*M1 B1 [4] | Uses integration of v(t) Award if +c omitted Allow +c or c omitted Accept 15.3, 46/3. Must be +ve | **Awarded if c omitted or assumed 0** |
| **ii** $t=0$ $x=0$ or $s=46/3$ hence x(0) or c = 0 or 46/3 Solves $t^2 – 9 = 0$ $t = (\pm)3$ $x(3) = 3'/3 -9x3 (+ 15.3)$ $x(3) = -18$ (or -2.67) Dist = 18 m | B1* M1 A1 D*M1 M1 D*B1 [6] | Needs explanation, may be seen in part i May be implied Value of t when direction of motion changes Substitutes cv(t) >2 in integrated x(t) Evaluates c – 18 may be implied award if … Accept 18(,0) [c=0 assumed] | **B1* awarded if limits 0 and 3 used correctly** Awarded if limits used correctly |
| **iii** $a = d(t^2 - 9)/dt$ $a = 2t$ 10 = 2t $t = 5$ $x(5) = (= 5^3/3 -9x5 + 15.3) = 12 \text{ m}$ OR $[t^3/3 - 9t]_3^5 = 12 \text{ m}$ | M1* A1 D*M1 A1 A1 [5] | Uses differentiation of v(t) |6 The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ of a particle at time $t \mathrm {~s}$ is given by $v = t ^ { 2 } - 9$. The particle travels in a straight line and passes through a fixed point $O$ when $t = 2$.\\
(i) Find the displacement of the particle from $O$ when $t = 0$.\\
(ii) Calculate the distance the particle travels from its position at $t = 0$ until it changes its direction of motion.\\
(iii) Calculate the distance of the particle from $O$ when the acceleration of the particle is $10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.
\hfill \mbox{\textit{OCR M1 2011 Q6 [15]}}