Challenging +1.2 This is a standard Further Maths mechanics question requiring integration of acceleration twice with boundary conditions, then elimination of the parameter t. While it involves multiple steps (integrate twice, apply two conditions at t=4, eliminate parameter), each step follows a routine procedure with no novel insight required. The algebra is straightforward and the method is textbook-standard for FM1/FM2 kinematics.
In this question you must show detailed reasoning.
In this question, positions are given relative to a fixed origin, O. The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are in the \(x\)- and \(y\)-directions respectively in a horizontal plane. Distances are measured in centimetres and the time, \(t\), is measured in seconds, where \(0 \leq t \leq 5\).
A small radio-controlled toy car C moves on a horizontal surface which contains O.
The acceleration of C is given by \(2\mathbf{i} + t\mathbf{j} \text{ cm s}^{-2}\).
When \(t = 4\), the displacement of C from O is \(16\mathbf{i} + \frac{32}{3}\mathbf{j}\) cm, and the velocity of C is \(8\mathbf{i} \text{ cm s}^{-1}\).
Determine a cartesian equation for the path of C for \(0 < t < 5\). You are not required to simplify your answer. [6]
At least one integration correct, but must have two components;
condone missing c, c′(and accept i, j form throughout) –
2t
condone v= +c(so the constant appearing as a scalar)
1t2
2
At least one integration correct, (follow-through their v) but
must have two components – condone missing d, d' - condone
constant appearing as a scalar
Must be substituting t= xinto a cubic equation/expression in t
– dependent on both previous M marks
Must be y = … or any equivalent un-simplified form containing
y (so must be an equation) – condone e.g.
y= 1( x)3−8( x)+96 isw
6 3
Question 6:
6 | 2t+c
(v=)
1t2 +c′
2
2t
(v=)
1t2 −8
2
t2 +d
(r=)
1t3−8t+d′
6
t2
(r=)
1t3−8t+32
6
Substitute t= x in equation for y
3 1
y= 1x2 −8x2 +32
6 | M1*
A1
M1dep*
A1
M1
A1
[6] | 3.4
1.1
1.1
1.1
2.5
1.1 | At least one integration correct, but must have two components;
condone missing c, c′(and accept i, j form throughout) –
2t
condone v= +c(so the constant appearing as a scalar)
1t2
2
At least one integration correct, (follow-through their v) but
must have two components – condone missing d, d' - condone
constant appearing as a scalar
Must be substituting t= xinto a cubic equation/expression in t
– dependent on both previous M marks
Must be y = … or any equivalent un-simplified form containing
y (so must be an equation) – condone e.g.
y= 1( x)3−8( x)+96 isw
6 3
\textbf{In this question you must show detailed reasoning.}
In this question, positions are given relative to a fixed origin, O. The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are in the $x$- and $y$-directions respectively in a horizontal plane. Distances are measured in centimetres and the time, $t$, is measured in seconds, where $0 \leq t \leq 5$.
A small radio-controlled toy car C moves on a horizontal surface which contains O.
The acceleration of C is given by $2\mathbf{i} + t\mathbf{j} \text{ cm s}^{-2}$.
When $t = 4$, the displacement of C from O is $16\mathbf{i} + \frac{32}{3}\mathbf{j}$ cm, and the velocity of C is $8\mathbf{i} \text{ cm s}^{-1}$.
Determine a cartesian equation for the path of C for $0 < t < 5$. You are \textbf{not} required to simplify your answer. [6]
\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2024 Q6 [6]}}