| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Piecewise motion functions |
| Difficulty | Standard +0.3 This is a straightforward multi-part kinematics question requiring standard differentiation and integration with given formulas. Part (i) is simple substitution into a derivative, part (ii) requires integrating acceleration to find velocity with an initial condition, and part (iii) requires integrating velocity to find displacement. All steps are routine M1 techniques with no problem-solving insight needed, making it slightly easier than average. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| 6(i) | \(v = 0.004t^3 - 0.12t^2 + 1.2t\) | M1 |
| 6(i) | \(v(10) = 4 - 12 + 12 = 4ms^{-1}\) | A1 |
| 6(i) | (AG) | |
| 6(i) | [3] | |
| 6(ii) | \(v = 0.8t - 0.04t^3\) (+C) | M1 |
| 6(ii) | \(8 - 4 + C = 4\) | A1 |
| 6(ii) | \(v = 0.8x20 - 0.04x20^2\) (+C) | M1* |
| 6(ii) | \(v(20) = 16 - 16 = 0\) | (AG) |
| 6(ii) | [5] | |
| 6(iii) | \(S = 0.4t^2 - 0.04t^3/3\) (+K) | M1 |
| 6(iii) | Accept 0.4t² – 0.013t³ (+ ct +K, must be linear) | A1 |
| 6(iii) | \(s(10) = 10 - 40 + 60 = 30\) | B1 |
| 6(iii) | \(40 - 40/3 + K = 30\) ⇒ K = 10/3 | M1 |
| 6(iii) | \(S(20) = 160 - 320/3 + 10/3 = 56.7m\) | A1 |
| 6(iii) | [6] | |
| 6(iii) | OR \(s(10) = 10 - 40 + 60 = 30\) | B1 |
| 6(iii) | \(S = 0.4t^2 - 0.04t^3/3\) | M1 |
| 6(iii) | Accept 0.4t² – 0.013t (+ ct +K, must be linear) | A1 |
| 6(iii) | \(S(20) - S(10) = 26.6, 26.7\) | A1 |
| 6(iii) | displacement is 56.7m | B1 |
6(i) | $v = 0.004t^3 - 0.12t^2 + 1.2t$ | M1 | For differentiating s
6(i) | $v(10) = 4 - 12 + 12 = 4ms^{-1}$ | A1 | Condone the inclusion of +c
6(i) | | (AG) | A1 | Correct formula for v (no +c) and t=10
6(i) | | [3] |
6(ii) | $v = 0.8t - 0.04t^3$ (+C) | M1 | For integrating a
6(ii) | $8 - 4 + C = 4$ | A1 |
6(ii) | $v = 0.8x20 - 0.04x20^2$ (+C) | M1* | Only for using v(10) = 4 to find C
6(ii) | $v(20) = 16 - 16 = 0$ | (AG) | DA1 | Dependant on M1*
6(ii) | | [5] |
6(iii) | $S = 0.4t^2 - 0.04t^3/3$ (+K) | M1 | For integrating v
6(iii) | Accept 0.4t² – 0.013t³ (+ ct +K, must be linear) | A1 |
6(iii) | $s(10) = 10 - 40 + 60 = 30$ | B1 |
6(iii) | $40 - 40/3 + K = 30$ ⇒ K = 10/3 | M1 | For using S(10) = 30 to find K
6(iii) | $S(20) = 160 - 320/3 + 10/3 = 56.7m$ | A1 | Not if S includes ct term
6(iii) | | [6] |
6(iii) | OR $s(10) = 10 - 40 + 60 = 30$ | B1 |
6(iii) | $S = 0.4t^2 - 0.04t^3/3$ | M1 | For integrating v
6(iii) | Accept 0.4t² – 0.013t (+ ct +K, must be linear) | A1 | Using limits of 10 and 20 (limits 0, 10 M0A0B0)
6(iii) | $S(20) - S(10) = 26.6, 26.7$ | A1 | For 53.3 - 26.7 or better (Note S(10) = 26.7 is fortuitously correct M0A0B0)
6(iii) | displacement is 56.7m | B1 | Accept 56.6 to 56.76 A particle starts from rest at the point A and travels in a straight line. The displacement sm of the particle from A at time ts after leaving A is given by
$$s = 0.001 t ^ { 4 } - 0.04 t ^ { 3 } + 0.6 t ^ { 2 } , \quad \text { for } 0 \leqslant t \leqslant 10 .$$
(i) Show that the velocity of the particle is $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when $\mathrm { t } = 10$.
The acceleration of the particle for $t \geqslant 10$ is $( 0.8 - 0.08 t ) \mathrm { m } \mathrm { s } ^ { - 2 }$.\\
(ii) Show that the velocity of the particle is zero when $\mathrm { t } = 20$.\\
(iii) Find the displacement from A of the particle when $\mathrm { t } = 20$.
\hfill \mbox{\textit{OCR M1 2007 Q6 [14]}}