6 A particle moves in a straight line $$\(A B\)$$. The velocity $$\(v \mathrm{~m} \mathrm{~s}^{-1}\)$$ of the particle $$\(t \mathrm{~s}\)$$ after leaving $$\(A\)$$ is given by $$\(v=k\left(t^{2}-10 t+21\right)\)$$, where $$\(k\)$$ is a constant. The displacement of the particle from $$\(A\)$$, in the direction towards $$\(B\)$$, is $$\(2.85 \mathrm{~m}\)$$ when $$\(t=3\)$$ and is $$\(2.4 \mathrm{~m}\)$$ when $$\(t=6\)$$. Find the value of $$\(k\)$$. Hence find an expression, in terms of $$\(t\)$$, for the displacement of the particle from $$\(A\)$$. ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mathematics
IGCSE&ALevel
CAIE
Exam No:9709_s20_qp_41 Year:2020 Question No:6(a)

Answer:

\(\int k\left(t^{2}-10 t+21\right) \mathrm{d} t\)
\(s=k\left(\frac{1}{3} t^{3}+5 t^{2}+21 t\right)+\mathrm{C}\)
\(2.85=k\left(\frac{1}{3} \times 3^{3}-5 \times 3^{2}+21 \times 3\right)+\mathrm{C}\) or \(2.4=k\left(\frac{1}{3} \times 6^{3}-5 \times 6^{2}+21 \times 6\right)+\mathrm{C}\)
$
2.85=27 k+\mathrm{C}, 2.4=18 k+\mathrm{C}
$

(A1 for both)
Solving for \(k\)
\(k=0.05\)
$
s=0.05\left(\frac{1}{3} t^{3}-5 t^{2}+21 t\right)+1.5
$

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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