CAIE M1 2021 June — Question 5 8 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionJune
Marks8
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Mark schemeDownload PDF ↗
TopicVariable acceleration (1D)
TypeVariable acceleration with initial conditions
DifficultyStandard +0.3 This is a straightforward variable acceleration problem requiring integration of a given acceleration function. Part (a) involves setting v=0 and solving a simple equation; part (b) requires integrating twice and evaluating at a specific time. The algebra is routine and the problem follows a standard template with no novel insight required, making it slightly easier than average.
Spec3.02f Non-uniform acceleration: using differentiation and integration
A particle moving in a straight line starts from rest at a point \(A\) and comes instantaneously to rest at a point \(B\). The acceleration of the particle at time \(t\) s after leaving \(A\) is \(a \text{ m s}^{-2}\), where $$a = 6t^{\frac{1}{2}} - 2t.$$
  1. Find the value of \(t\) at point \(B\). [3]
  2. Find the distance travelled from \(A\) to the point at which the acceleration of the particle is again zero. [5]
Question 5:
AnswerMarks
5Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
AnswerMarks
5(a) 1 
v=6t2 −2tdt
 
AnswerMarks Guidance
 M1 For integration. v=at is M0.
3
AnswerMarks Guidance
v=4t2 −t2 (+c )A1 Allow unsimplified coefficients.
1
AnswerMarks
v = 0 leading to t = 0 or t2 =4 leading to t = 16A1
3
AnswerMarks Guidance
5(b)1
6t2 −2t =0M1 Attempt to solve a = 0, using valid algebra, reaching t = …
t = 9A1
 3 
s=4t2 −t2dt
 
 
 s= 8 t 5 2 − 1 t3 (+c )  
AnswerMarks Guidance
 5 3 M1 For integration of their expression for v which includes a term with
a fractional power. Allow unsimplified coefficients. v=at is M0
8 5 1
s = t2 − t3
AnswerMarks Guidance
5 3A1 For correct integral
Distance = 145.8 mB1 729
Allow or 146 to 3s.f.
5
5
AnswerMarks Guidance
QuestionAnswer Marks 
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) |  1 
v=6t2 −2tdt
 
  | M1 | For integration. v=at is M0.
3
v=4t2 −t2 (+c ) | A1 | Allow unsimplified coefficients.
1
v = 0 leading to t = 0 or t2 =4 leading to t = 16 | A1
3
--- 5(b) ---
5(b) | 1
6t2 −2t =0 | M1 | Attempt to solve a = 0, using valid algebra, reaching t = …
t = 9 | A1
 3 
s=4t2 −t2dt
 
 
 s= 8 t 5 2 − 1 t3 (+c )  
 5 3  | M1 | For integration of their expression for v which includes a term with
a fractional power. Allow unsimplified coefficients. v=at is M0
8 5 1
s = t2 − t3
5 3 | A1 | For correct integral
Distance = 145.8 m | B1 | 729
Allow or 146 to 3s.f.
5
5
Question | Answer | Marks | Guidance
A particle moving in a straight line starts from rest at a point $A$ and comes instantaneously to rest at a point $B$. The acceleration of the particle at time $t$ s after leaving $A$ is $a \text{ m s}^{-2}$, where
$$a = 6t^{\frac{1}{2}} - 2t.$$

\begin{enumerate}[label=(\alph*)]
\item Find the value of $t$ at point $B$. [3]
\item Find the distance travelled from $A$ to the point at which the acceleration of the particle is again zero. [5]
\end{enumerate}

\hfill \mbox{\textit{CAIE M1 2021 Q5 [8]}}
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