| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | SUVAT single equation: straightforward find |
| Difficulty | Standard +0.3 This is a straightforward SUVAT problem followed by basic calculus application. Part (i) uses standard constant acceleration equations. Part (ii) involves differentiating v = kt³ to find acceleration and using given conditions to find k. All steps are routine with clear signposting, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([1.3 = 0.9 + 0.004T, 1.3^2 = 0.9^2 + 2 \times 0.004S]\) | M1 | For using \(v = u + at\) or \(v^2 = u^2 + 2as\) |
| Time is 100 s (or distance is 110 m) | A1 | |
| Distance is 110 m (or time is 100 s) | B1 | [3] |
| (ii) \(\int kr^7 dt = \frac{1}{4}kt^4\) | B1 | |
| \([k(\frac{1}{4} \times 100^4 - 0) = 110]\) | M1 | For using limits 0 to T and equating definite integral to S |
| \(k = 4.4 \times 10^{-6}\) | A1 | |
| \([v_W = 0.9 + 0.004 \times 64.05, v_C = 4.4 \times 10^{-6} \times 64.05^4]\) | M1 | For attempting to find the speed of the walker and of the cyclist. |
| Both are equal to 1.16 ms \(^{-1}\) correct to 3 sf. | A1 | [5] |
| (iii) Acceleration \(= 3kt^2\) | B1 | |
| Acceleration at B is 0.132 ms\(^{-2}\) | B1 | [2] |
**(i)** $[1.3 = 0.9 + 0.004T, 1.3^2 = 0.9^2 + 2 \times 0.004S]$ | M1 | For using $v = u + at$ or $v^2 = u^2 + 2as$
Time is 100 s (or distance is 110 m) | A1 |
Distance is 110 m (or time is 100 s) | B1 | [3]
**(ii)** $\int kr^7 dt = \frac{1}{4}kt^4$ | B1 |
$[k(\frac{1}{4} \times 100^4 - 0) = 110]$ | M1 | For using limits 0 to T and equating definite integral to S
$k = 4.4 \times 10^{-6}$ | A1 |
$[v_W = 0.9 + 0.004 \times 64.05, v_C = 4.4 \times 10^{-6} \times 64.05^4]$ | M1 | For attempting to find the speed of the walker and of the cyclist.
Both are equal to 1.16 ms $^{-1}$ correct to 3 sf. | A1 | [5]
**(iii)** Acceleration $= 3kt^2$ | B1 |
Acceleration at B is 0.132 ms$^{-2}$ | B1 | [2]7 A walker travels along a straight road passing through the points $A$ and $B$ on the road with speeds $0.9 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and $1.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ respectively. The walker's acceleration between $A$ and $B$ is constant and equal to $0.004 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
(i) Find the time taken by the walker to travel from $A$ to $B$, and find the distance $A B$.
A cyclist leaves $A$ at the same instant as the walker. She starts from rest and travels along the straight road, passing through $B$ at the same instant as the walker. At time $t \mathrm {~s}$ after leaving $A$ the cyclist's speed is $k t ^ { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$, where $k$ is a constant.\\
(ii) Show that when $t = 64.05$ the speed of the walker and the speed of the cyclist are the same, correct to 3 significant figures.\\
(ii) Find the cyclist's acceleration at the instant she passes through $B$.
\hfill \mbox{\textit{CAIE M1 2011 Q7 [10]}}