Standard +0.3 This is a straightforward circular motion problem requiring recall of standard formulas: radial acceleration = v²/r and transverse acceleration = dv/dt. Students substitute the given velocity function, use the radial acceleration condition to find T, then differentiate to find the transverse component. While it involves multiple steps, each is routine application of memorized formulas with no conceptual difficulty or novel insight required.
A particle \(P\) is moving in a circle of radius 2 m. At time \(t\) seconds, its velocity is \((t - 1)^2\) m s\(^{-1}\). At a particular time \(T\) seconds, where \(T > 0\), the magnitude of the radial component of the acceleration of \(P\) is 8 m s\(^{-2}\). Find the magnitude of the transverse component of the acceleration of \(P\) at this instant.
[5]
Find magnitude of transverse acceleration at t = T
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Answer
Marks
Guidance
Question
Answer
Marks
Question 1:
1 | (T – 1)4 / 2 = 8 | M1 A1 | Equate radial acceln. to 8 at t = T from v2/r
T = 3 (or T – 1 = 2) | A1 | Hence find positive value of T (or of T – 1)
a = 2 (T – 1) = 4 [m s-2]
T | M1 A1 | Find magnitude of transverse acceleration at t = T
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Question | Answer | Marks | Guidance
A particle $P$ is moving in a circle of radius 2 m. At time $t$ seconds, its velocity is $(t - 1)^2$ m s$^{-1}$. At a particular time $T$ seconds, where $T > 0$, the magnitude of the radial component of the acceleration of $P$ is 8 m s$^{-2}$. Find the magnitude of the transverse component of the acceleration of $P$ at this instant.
[5]
\hfill \mbox{\textit{CAIE FP2 2019 Q1 [5]}}