Standard +0.3 This is a straightforward application of Newton's second law (F = ma) with vectors. Students must find the force vector by multiplying mass by acceleration, then use the magnitude formula to set up and solve a simple quadratic equation. All steps are routine with no conceptual challenges beyond basic vector operations.
9 In this question the horizontal unit vectors \(\mathbf { i }\) and \(\mathbf { j }\) are in the directions east and north respectively.
A model ship of mass 2 kg is moving so that its acceleration vector \(\mathbf { a m s } ^ { - 2 }\) at time \(t\) seconds is given by \(\mathbf { a } = 3 ( 2 t - 5 ) \mathbf { i } + 4 \mathbf { j }\). When \(t = T\), the magnitude of the horizontal force acting on the ship is 10 N .
Find the possible values of \(T\).
## Question 9:
$F = \sqrt{36(2T-5)^2 + 64}$ | **M1\*** | AO3.3 | Correct use of $\mathbf{F} = m\mathbf{a}$ and Pythagoras
$36(2T-5)^2 = 36$ | **A1** | AO1.1 | Correct equation(s) for both values of $T$; e.g. $10 = 2\sqrt{9(2T-5)^2 + 16}$ | Allow $t$ throughout
$2T - 5 = \pm 1 \Rightarrow T = ...$ | **Dep\*M1** | AO1.1 | Attempt to solve a quadratic leading to at least one value for $T$
$T = 2$ and $T = 3$ | **A1** | AO2.2a |
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9 In this question the horizontal unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the directions east and north respectively.\\
A model ship of mass 2 kg is moving so that its acceleration vector $\mathbf { a m s } ^ { - 2 }$ at time $t$ seconds is given by $\mathbf { a } = 3 ( 2 t - 5 ) \mathbf { i } + 4 \mathbf { j }$. When $t = T$, the magnitude of the horizontal force acting on the ship is 10 N .
Find the possible values of $T$.
\hfill \mbox{\textit{OCR PURE Q9 [4]}}