Standard +0.3 This is a straightforward two-step integration problem: integrate acceleration to find velocity (using initial condition v=1.8), then integrate velocity to find displacement at t=16. The power of t^{-0.75} is non-standard but the integration is mechanical using the standard power rule, making this slightly easier than average.
3 A particle \(P\) moves in a straight line. It starts from a point \(O\) on the line with velocity \(1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). The acceleration of \(P\) at time \(t \mathrm {~s}\) after leaving \(O\) is \(0.8 t ^ { - 0.75 } \mathrm {~m} \mathrm {~s} ^ { - 2 }\). Find the displacement of \(P\) from \(O\) when \(t = 16\).
| M1 | For using $v = \int a \, dt$
$v = (0.8/0.25)t^{0.25} + (C)$ | A1 |
$C = 1.8$ | B1 |
| M1 | For using $s = \int v \, dt$
$s = (3.2/1.25)t^{1.25} + 1.8t + (K)$ | A1 ft | ft only from an incorrect non-zero value of $C$
Distance is $111 \text{ m}$ | A1 | 6
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3 A particle $P$ moves in a straight line. It starts from a point $O$ on the line with velocity $1.8 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. The acceleration of $P$ at time $t \mathrm {~s}$ after leaving $O$ is $0.8 t ^ { - 0.75 } \mathrm {~m} \mathrm {~s} ^ { - 2 }$. Find the displacement of $P$ from $O$ when $t = 16$.
\hfill \mbox{\textit{CAIE M1 2011 Q3 [6]}}