| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Multi-phase journey: find unknown speed or time |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question involving piecewise motion analysis and basic calculus (differentiation of polynomials). Part (i) requires setting up equations from a velocity-time graph, parts (ii-iii) involve straightforward differentiation, and part (iv) requires substitution. All techniques are routine for M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_{\text{dec}} = 15 \times 4 - 1.75 \times 4^2/2\) | M1 | Or \(v = 15 - 1.75 \times 4\) and \(s = (15+v)/2 \times 4\) |
| \(S_{\text{dec}} = 46\) | A1 | May be implied |
| \(100 - 46 = 15T/2 + 15(10-4-T) \quad (= 15 \times 6 - 15T/2)\) | M1 | Any attempt at combined 3 stage distances being 100 |
| \(54 = 90 - 7.5T\) | A1 ft | Simplification not essential. ft \(\text{cv}(S_{\text{dec}}(\mathbf{i}))\), numerical |
| \(T = 4.8\) | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(V_R = d(3t^2 - 0.2t^3)/dt\) | M1 | Attempt at differentiating \(S_R\) |
| \(V_R = 6t - 0.6t^2\) | A1 | Accept \(V_R = 2 \times 3t - 3 \times 0.2t^2\) |
| \(V_R(5) (= 6 \times 5 - 0.6 \times 5^2) = 15 \text{ ms}^{-1}\) | A1 [3] AG | Must show explicit substitution |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A_R = d(6t - 0.6t^2)/dt\) | M1* | Attempt at differentiating \(V_R\) |
| \(6 - 1.2t = -1.75\) | D*M1 | Must be \(-1.75\) or \(1.2t - 6 = 1.75\) (i.e. employs deceleration) |
| \(t = 6.46\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(S_R(10) = 3 \times 10^2 - 0.2 \times 10^3\) | M1 | Substitutes 10 into \(S_R\) formula |
| \(S_R(10) = 100\) | A1 [2] | |
| OR \(3t^2 - 0.2t^3 = 100\), \(t = 10\) which is how long the athlete takes to finish | M1, A1 | Sets up and tries to solve equation for robot. Needs comment about athlete or both finishing race in 10 s |
# Question 5:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_{\text{dec}} = 15 \times 4 - 1.75 \times 4^2/2$ | M1 | Or $v = 15 - 1.75 \times 4$ and $s = (15+v)/2 \times 4$ |
| $S_{\text{dec}} = 46$ | A1 | May be implied |
| $100 - 46 = 15T/2 + 15(10-4-T) \quad (= 15 \times 6 - 15T/2)$ | M1 | Any attempt at combined 3 stage distances being 100 |
| $54 = 90 - 7.5T$ | A1 ft | Simplification not essential. ft $\text{cv}(S_{\text{dec}}(\mathbf{i}))$, numerical |
| $T = 4.8$ | A1 **[5]** | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $V_R = d(3t^2 - 0.2t^3)/dt$ | M1 | Attempt at differentiating $S_R$ |
| $V_R = 6t - 0.6t^2$ | A1 | Accept $V_R = 2 \times 3t - 3 \times 0.2t^2$ |
| $V_R(5) (= 6 \times 5 - 0.6 \times 5^2) = 15 \text{ ms}^{-1}$ | A1 **[3]** AG | Must show explicit substitution |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A_R = d(6t - 0.6t^2)/dt$ | M1* | Attempt at differentiating $V_R$ |
| $6 - 1.2t = -1.75$ | D*M1 | Must be $-1.75$ or $1.2t - 6 = 1.75$ (i.e. employs deceleration) |
| $t = 6.46$ | A1 **[3]** | |
## Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $S_R(10) = 3 \times 10^2 - 0.2 \times 10^3$ | M1 | Substitutes 10 into $S_R$ formula |
| $S_R(10) = 100$ | A1 **[2]** | |
| OR $3t^2 - 0.2t^3 = 100$, $t = 10$ which is how long the athlete takes to finish | M1, A1 | Sets up and tries to solve equation for robot. Needs comment about athlete or both finishing race in 10 s |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{2b3457b6-1fe9-4e67-91d4-a8bc4a5b1709-3_394_789_251_639}
The diagram shows the ( $t , v$ ) graph of an athlete running in a straight line on a horizontal track in a 100 m race. He starts from rest and has constant acceleration until he reaches a speed of $15 \mathrm {~ms} ^ { - 1 }$ when $t = T$. He maintains this constant speed until he decelerates at a constant rate of $1.75 \mathrm {~ms} ^ { - 2 }$ for the final 4 s of the race. He completes the race in 10 s .\\
(i) Calculate $T$.
The athlete races against a robot which has a displacement from the starting line of $\left( 3 t ^ { 2 } - 0.2 t ^ { 3 } \right) \mathrm { m }$, at time $t \mathrm {~s}$ after the start of the race.\\
(ii) Show that the speed of the robot is $15 \mathrm {~ms} ^ { - 1 }$ when $t = 5$.\\
(iii) Find the value of $t$ for which the decelerations of the robot and the athlete are equal.\\
(iv) Verify that the athlete and the robot reach the finish line simultaneously.
\hfill \mbox{\textit{OCR M1 2012 Q5 [13]}}