Moderate -0.8 This is a straightforward kinematics integration problem requiring two integrations with given initial conditions. While it requires careful algebraic manipulation to reach the specific form requested, the techniques are standard AS-level mechanics with no conceptual difficulty or novel problem-solving required. The 'show that' format provides the target answer, making it easier than an open-ended question.
A particle moves in a straight line with acceleration \(a\) m s\(^{-2}\), at time \(t\) seconds, where
$$a = 10 - 6t$$
The particle's velocity, \(v\) m s\(^{-1}\), and displacement, \(r\) metres, are both initially zero.
Show that
$$r = t^2(5 - t)$$
Fully justify your answer.
[4 marks]
Question 17:
17 | Integrates to find v with at least
one non-constant term correct | 3.4 | M1 | v=∫10−6t dt
v=10t−3t2 +c
When t = 0, v = 0 therefore c = 0
r =∫10t−3t2 dt
r =5t2 −t3 +c
When t = 0, r = 0 therefore c = 0
So
r =t2(5−t)
Obtains fully correct expression
for v
Must include +c or state that c =
0 | 1.1b | A1
Integrates their v with at least
one term correct | 1.1a | M1
Completes reasoned argument
to show r =t2(5−t)
Explanation for both constants
of integration = 0 must be given | 2.1 | R1
Question 17 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
A particle moves in a straight line with acceleration $a$ m s$^{-2}$, at time $t$ seconds, where
$$a = 10 - 6t$$
The particle's velocity, $v$ m s$^{-1}$, and displacement, $r$ metres, are both initially zero.
Show that
$$r = t^2(5 - t)$$
Fully justify your answer.
[4 marks]
\hfill \mbox{\textit{AQA AS Paper 1 2024 Q17 [4]}}