| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Total distance with direction changes |
| Difficulty | Standard +0.3 This is a straightforward variable acceleration question requiring standard calculus techniques: solving a cubic equation (factorizable), finding maximum acceleration via differentiation, and computing distance with definite integration. All steps are routine A-level mechanics procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.07n Stationary points: find maxima, minima using derivatives1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| 7(i) | –0.01t(t2 – 22t + 40) = 0 | |
| –0.01t(t – 20)(t – 2) = 0 | M1 | Attempting to solve v = 0 for t for a |
| Answer | Marks |
|---|---|
| t = 2 or t = 20 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(ii) | a = – 0.03t2 + 0.44t – 0.4 | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 0.44 – 0.06t = 0 | M1 | For differentiation or finding values of |
| Answer | Marks | Guidance |
|---|---|---|
| Max acceleration when t = 7.33 | A1 | 22 |
| Answer | Marks | Guidance |
|---|---|---|
| 7(iii) | ( ) | |
| ∫ −0.01t3 +0.22t2 −0.4t dt | *M1 | For using integration. |
| Answer | Marks | Guidance |
|---|---|---|
| 4 3 | A1 | Correct Integration |
| Answer | Marks | Guidance |
|---|---|---|
| s 20 −s 2 | DM1 | Limits 2 and 20 used correctly |
| Answer | Marks | Guidance |
|---|---|---|
| Distance = 107 m | A1 | 2673 |
Question 7:
--- 7(i) ---
7(i) | –0.01t(t2 – 22t + 40) = 0
–0.01t(t – 20)(t – 2) = 0 | M1 | Attempting to solve v = 0 for t for a
solvable quadratic using factors or
quadratic formula and obtaining two non-
zero solutions
t = 2 or t = 20 | A1
2
--- 7(ii) ---
7(ii) | a = – 0.03t2 + 0.44t – 0.4 | M1 | For differentiation
a is greatest (maximum) when
0.44 – 0.06t = 0 | M1 | For differentiation or finding values of
t = t and t = t where a = 0 and using
1 2
t = ½(t + t ) or completing the square or
1 2
other method to find maximum value
Max acceleration when t = 7.33 | A1 | 22
Allow t =
3
3
--- 7(iii) ---
7(iii) | ( )
∫ −0.01t3 +0.22t2 −0.4t dt | *M1 | For using integration.
s ( t )=− 0.01 t4 + 0.22 t3 −0.2t2
4 3 | A1 | Correct Integration
Allow + C included
( ) ( )
s 20 −s 2 | DM1 | Limits 2 and 20 used correctly
Dependent on previous M1 having been
scored
Distance = 107 m | A1 | 2673
Distance = = 106.92
25
4A particle starts from rest and moves in a straight line. The velocity of the particle at time $t$ s after the start is $v$ m s$^{-1}$, where
$$v = -0.01t^3 + 0.22t^2 - 0.4t.$$
\begin{enumerate}[label=(\roman*)]
\item Find the two positive values of $t$ for which the particle is instantaneously at rest. [2]
\item Find the time at which the acceleration of the particle is greatest. [3]
\item Find the distance travelled by the particle while its velocity is positive. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q7 [9]}}