| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Variable acceleration with initial conditions |
| Difficulty | Standard +0.3 This is a straightforward integration problem requiring students to integrate acceleration to find velocity, solve for time, then integrate velocity to find distance. While it involves non-constant acceleration (making it slightly above basic kinematics), the integration is simple polynomial form and the problem follows a standard two-step procedure with clear given values and no conceptual tricks. |
| Spec | 1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For integrating \(a(t)\) to obtain \(v(t)\) | |
| \(V(t) = 1.5t + 0.006t^2\) | A1 | Constant of integration zero or absent |
| \([0.006t^2 + 1.5t - 90 = 0 \Rightarrow t^2 + 250t - 15000 = 0 \Rightarrow (t-50)(t+300) = 0]\) | DM1 | For using \(v(t) = 90\) and solving for \(t\) (dependent on integration) |
| Leaves the ground when \(t = 50\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For integrating \(v(t)\) and using limits 0 to candidate's answer for part (i) | |
| \(s = 0.75t^2 + 0.002t^3\) | A1ft | ft if there is a non-zero constant of integration \(C\) in part (i): \(s = 0.75t^2 + 0.002t^3 + Ct\) |
| Distance is \(2125\ \text{m}\) | A1ft [3] | Accept 2120 or 2130; ft \(t\) from part (i) in \(0.75t^2 + 0.002t^3\) |
## Question 4:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For integrating $a(t)$ to obtain $v(t)$ |
| $V(t) = 1.5t + 0.006t^2$ | A1 | Constant of integration zero or absent |
| $[0.006t^2 + 1.5t - 90 = 0 \Rightarrow t^2 + 250t - 15000 = 0 \Rightarrow (t-50)(t+300) = 0]$ | DM1 | For using $v(t) = 90$ and solving for $t$ (dependent on integration) |
| Leaves the ground when $t = 50$ | A1 [4] | |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For integrating $v(t)$ and using limits 0 to candidate's answer for part (i) |
| $s = 0.75t^2 + 0.002t^3$ | A1ft | ft if there is a non-zero constant of integration $C$ in part (i): $s = 0.75t^2 + 0.002t^3 + Ct$ |
| Distance is $2125\ \text{m}$ | A1ft [3] | Accept 2120 or 2130; ft $t$ from part (i) in $0.75t^2 + 0.002t^3$ |
---4 An aeroplane moves along a straight horizontal runway before taking off. It starts from rest at $O$ and has speed $90 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at the instant it takes off. While the aeroplane is on the runway at time $t$ seconds after leaving $O$, its acceleration is $( 1.5 + 0.012 t ) \mathrm { m } \mathrm { s } ^ { - 2 }$. Find\\
(i) the value of $t$ at the instant the aeroplane takes off,\\
(ii) the distance travelled by the aeroplane on the runway.
\hfill \mbox{\textit{CAIE M1 2013 Q4 [7]}}