A particle travels in a straight line $$\(P Q\)$$. The velocity of the particle $$\(t \mathrm{~s}\)$$ after leaving $$\(P\)$$ is $$\(v \mathrm{~m} \mathrm{~s}^{-1}\)$$, where $$\[ v=4.5+4 t-0.5 t^{2} . \]$$ The particle comes to instantaneous rest at Q Find the distance $$\(P Q\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................
Exam No：9709_s20_qp_43 Year：2020 Question No：6(b)
Answer:
Velocity \(=0\) when \(4.5+4 t-0.5 t^{2}=0\)
\(t=9(\operatorname{reject} t=-1)\)
\(\int\left(4.5+4 t-0.5 t^{2}\right) d t\)
\(4.5 t+2 t^{2}-\frac{1}{6} t^{3}[+c]\)
Apply limits ( 0 and 9\()\)
Distance \(=81 \mathrm{~m}\)
\(t=9(\operatorname{reject} t=-1)\)
\(\int\left(4.5+4 t-0.5 t^{2}\right) d t\)
\(4.5 t+2 t^{2}-\frac{1}{6} t^{3}[+c]\)
Apply limits ( 0 and 9\()\)
Distance \(=81 \mathrm{~m}\)
Knowledge points:
4.2.3 use differentiation and integration with respect to time to solve simple problems concerning displacement, velocity and acceleration Calculus required is restricted to techniques from the content for Paper 1: Pure Mathematics 1.
Solution:
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