CAIE M1 2020 Specimen — Question 6 11 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2020
SessionSpecimen
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypePiecewise motion functions
DifficultyModerate -0.3 This is a standard piecewise velocity function question requiring continuity to find k, sketching, identifying where acceleration is positive (v' > 0), and integration for distance. All techniques are routine M1 material with straightforward calculus, making it slightly easier than average A-level.
Spec3.02f Non-uniform acceleration: using differentiation and integration3.02g Two-dimensional variable acceleration

6 A particle \(P\) moves in a straight line. The velocity \(v \mathrm {~m} \mathrm {~s} ^ { - 1 }\) at time \(t \mathrm {~s}\) is given by $$\begin{array} { l l } v = 5 t ( t - 2 ) & \text { for } 0 \leqslant t \leqslant 4 \\ v = k & \text { for } 4 \leqslant t \leqslant 14 \\ v = 68 - 2 t & \text { for } 14 \leqslant t \leqslant 20 \end{array}$$ where \(k\) is a constant.
  1. Find \(k\).
  2. Sketch the velocity-time graph for \(0 \leqslant t \leqslant 20\).
  3. Find the set of values of \(t\) for which the acceleration of \(P\) is positive.
  4. Find the total distance travelled by \(P\) in the interval \(0 \leqslant t \leqslant 20\).

AnswerMarks Guidance
6(a)\(k = 40\) B1
6(b)Correct for \(0 \leq x < 4\) For horizontal line at \(y = 20\); for \(4 \leq x \leq 20\) (20, 28); \(y = 4\); \(y = 14\) to \(y = 20\) BFT BFT BFT
6(c)For attempting to differentiate to find \(a = 10 - 10\) Allow < 4 but not 1 M1 A1
6(d)\(\text{Dist} = \left[\frac{5t^2}{s^2} - \frac{100t}{3} + \frac{5 \times (40-10)}{2}\right] + (40 - 10) + (0.5 \times (40 + 28) \times 6)\) \(= 64\text{m}\) A1
**6(a)** | $k = 40$ | B1 | 

**6(b)** | Correct for $0 \leq x < 4$ For horizontal line at $y = 20$; for $4 \leq x \leq 20$ (20, 28); $y = 4$; $y = 14$ to $y = 20$ | BFT BFT BFT | Correct for 10 ≤ $x$ ≤ 14; FT on $k$

**6(c)** | For attempting to differentiate to find $a = 10 - 10$ Allow < 4 but not 1 | M1 A1 | For correct integration

**6(d)** | $\text{Dist} = \left[\frac{5t^2}{s^2} - \frac{100t}{3} + \frac{5 \times (40-10)}{2}\right] + (40 - 10) + (0.5 \times (40 + 28) \times 6)$ $= 64\text{m}$ | A1 | 

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6 A particle $P$ moves in a straight line. The velocity $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ at time $t \mathrm {~s}$ is given by

$$\begin{array} { l l } 
v = 5 t ( t - 2 ) & \text { for } 0 \leqslant t \leqslant 4 \\
v = k & \text { for } 4 \leqslant t \leqslant 14 \\
v = 68 - 2 t & \text { for } 14 \leqslant t \leqslant 20
\end{array}$$

where $k$ is a constant.\\
(a) Find $k$.\\
(b) Sketch the velocity-time graph for $0 \leqslant t \leqslant 20$.\\
(c) Find the set of values of $t$ for which the acceleration of $P$ is positive.\\
(d) Find the total distance travelled by $P$ in the interval $0 \leqslant t \leqslant 20$.\\

\hfill \mbox{\textit{CAIE M1 2020 Q6 [11]}}