1 A particle \(P\) is moving in a circle of radius 0.25 m . At time \(t\) seconds, its velocity is \(\left( 2 t ^ { 2 } - 4 t + 3 \right) \mathrm { m } \mathrm { s } ^ { - 1 }\). At a particular time \(T\) seconds, where \(T > 0\), the magnitude of the transverse component of the acceleration of \(P\) is \(6 \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the magnitude of the radial component of the acceleration of \(P\) at this instant.

## Question 1:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $4T - 4 = 6$, $T = 2.5$ | M1 A1 | Find $T$ by equating $dv/dt$ at $t=T$ to 6 |
| $v^2/r = (2T^2 - 4T + 3)^2/0.25$ | | Find radial component $v^2/r$ of acceleration at $t=T$ |
| $= (11/2)^2 \times 4 = 121$ [m s$^{-2}$] | M1 A1 | M0 if $T$ not given a value |
| **Total: 4 marks** | | SR: Max M1 (1/4) if linear and angular confused |

1 A particle $P$ is moving in a circle of radius 0.25 m . At time $t$ seconds, its velocity is $\left( 2 t ^ { 2 } - 4 t + 3 \right) \mathrm { m } \mathrm { s } ^ { - 1 }$. At a particular time $T$ seconds, where $T > 0$, the magnitude of the transverse component of the acceleration of $P$ is $6 \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the magnitude of the radial component of the acceleration of $P$ at this instant.

\hfill \mbox{\textit{CAIE FP2 2015 Q1 [4]}}