WJEC Unit 2 2018 June — Question 07 3 marks

Exam BoardWJEC
ModuleUnit 2 (Unit 2)
Year2018
SessionJune
Marks3
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TopicVariable acceleration (1D)
TypeDisplacement from velocity by integration
DifficultyModerate -0.8 This is a straightforward integration question requiring students to integrate a quadratic velocity function and use an initial condition to find the constant. It's a standard mechanics/calculus exercise with clear steps: integrate v to get s, substitute t=1 and s=-4 to find C, making it easier than average but not trivial.
Spec1.08a Fundamental theorem of calculus: integration as reverse of differentiation3.02f Non-uniform acceleration: using differentiation and integration

A particle moves along the horizontal \(x\)-axis so that its velocity \(v\) ms\(^{-1}\) at time \(t\) seconds is given by $$v = 6t^2 - 8t - 5.$$ At time \(t = 1\), the particle's displacement from the origin is \(-4\) m. Find an expression for the displacement of the particle at time \(t\) seconds. [3]

A particle moves along the horizontal $x$-axis so that its velocity $v$ ms$^{-1}$ at time $t$ seconds is given by
$$v = 6t^2 - 8t - 5.$$

At time $t = 1$, the particle's displacement from the origin is $-4$ m. Find an expression for the displacement of the particle at time $t$ seconds. [3]

\hfill \mbox{\textit{WJEC Unit 2 2018 Q07 [3]}}