Question 12:
Solution by graph sketching is not accepted.
Two particles, A and B, move along a straight line. The velocity of particle A, vA m s-1, is given by vA = 24t – 4t2 – 20, while the velocity of particle B, vB m s-1, is given by vB = 24t – 25, such that t is time, in seconds, after both particles passing fixed point O.
Calculate
(a) the maximum velocity, in m s-1, of particle A, [3 marks]
(b)(i) the time, in seconds, when particle A and particle B meet,
(ii) the total distance, in m , travelled by particle A in the first 5 seconds.
[7 marks]
Answer:
(a) $$ \begin{aligned} v_{\mathrm{A}}=24 t & -4 t^2-20 \\ \frac{\mathrm{~d} v}{\mathrm{~d} t} & =0 \\ 24-8 t & =0 \\ t & =3 \mathrm{~s} \end{aligned} $$
$$ \begin{aligned} v_{\text { maximum }} & =24(3)-4(3)^2-20 \\ & =16 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b) $$ \begin{aligned} s_{\mathrm{A}} & =\int\left(24 t-4 t^2-20\right) \mathrm{d} t \\ & =\frac{24 t^2}{2}-\frac{4 t^3}{3}-20 t+c \end{aligned} $$
$$ \begin{aligned} & t=0, s_{\mathrm{A}}=0 \\ & 0=12(0)^2-\frac{4}{3}(0)^3-20(0)+c \\ & c=0 \\ & s_{\mathrm{A}}=12 t^2-\frac{4}{3} t^3-20 t \end{aligned} $$
$$ \begin{aligned} s_{\mathrm{B}} & =\int(24 t-25) \mathrm{d} t \\ & =\frac{24 t^2}{2}-25 t+c \end{aligned} $$
$$ \begin{aligned} & t=0, s_{\mathrm{B}}=0 \\ & 0=12(0)^2-25(0)+c \\ & c=0 \\ & s_{\mathrm{B}}=12 t^2-25 t \end{aligned} $$
$$ \begin{aligned} S_{\mathrm{A}} & =S_{\mathrm{B}} \\ 12 t^2-\frac{4}{3} t^3-20 t & =12 t^2-25 t \\ -\frac{4}{3} t^3+5 t & =0 \\ -t\left(\frac{4}{3} t^2-5\right) & =0 \\ \frac{4}{3} t^2 & =5 \\ t & =\sqrt{\frac{15}{4}} \\ & =1.936 \mathrm{~s} \end{aligned} $$
(c) $$ \begin{aligned} & v_{\mathrm{A}}=0 \\ & 24 t-4 t^2-20=0 \\ & t^2-6 t+5=0 \\ & (t-1)(t-5)=0 \\ & t=1 \mathrm{~s}, 5 \mathrm{~s} \end{aligned} $$
$$ \begin{aligned} s & =\left|\int_0^1\left(24 t-4 t^2-20\right) \mathrm{d} t\right|+\int_1^5\left(24 t-4 t^2-20\right) \mathrm{d} t \\ & =\left|\left[12 t^2-\frac{4}{3} t^3-20 t\right]_0^1\right|+\left[12 t^2-\frac{4}{3} t^3-20 t\right]_1^5 \\ & =\left|\left[12(1)^2-\frac{4}{3}(1)^3-20(1)\right]-\left[12(0)^2-\frac{4}{3}(0)^3-20(0)\right]\right|+ \\ & \quad\left[\left[12(5)^2-\frac{4}{3}(5)^3-20(5)\right]-\left[12(1)^2-\frac{4}{3}(1)^3-20(1)\right]\right] \\ & =\frac{28}{3}+\frac{128}{3} \\ & =52 \mathrm{~m} \end{aligned} $$
Solution by graph sketching is not accepted.
Two particles, A and B, move along a straight line. The velocity of particle A, vA m s-1, is given by vA = 24t – 4t2 – 20, while the velocity of particle B, vB m s-1, is given by vB = 24t – 25, such that t is time, in seconds, after both particles passing fixed point O.
Calculate
(a) the maximum velocity, in m s-1, of particle A, [3 marks]
(b)(i) the time, in seconds, when particle A and particle B meet,
(ii) the total distance, in m , travelled by particle A in the first 5 seconds.
[7 marks]
Answer:
(a) $$ \begin{aligned} v_{\mathrm{A}}=24 t & -4 t^2-20 \\ \frac{\mathrm{~d} v}{\mathrm{~d} t} & =0 \\ 24-8 t & =0 \\ t & =3 \mathrm{~s} \end{aligned} $$
$$ \begin{aligned} v_{\text { maximum }} & =24(3)-4(3)^2-20 \\ & =16 \mathrm{~m} \mathrm{~s}^{-1} \end{aligned} $$
(b) $$ \begin{aligned} s_{\mathrm{A}} & =\int\left(24 t-4 t^2-20\right) \mathrm{d} t \\ & =\frac{24 t^2}{2}-\frac{4 t^3}{3}-20 t+c \end{aligned} $$
$$ \begin{aligned} & t=0, s_{\mathrm{A}}=0 \\ & 0=12(0)^2-\frac{4}{3}(0)^3-20(0)+c \\ & c=0 \\ & s_{\mathrm{A}}=12 t^2-\frac{4}{3} t^3-20 t \end{aligned} $$
$$ \begin{aligned} s_{\mathrm{B}} & =\int(24 t-25) \mathrm{d} t \\ & =\frac{24 t^2}{2}-25 t+c \end{aligned} $$
$$ \begin{aligned} & t=0, s_{\mathrm{B}}=0 \\ & 0=12(0)^2-25(0)+c \\ & c=0 \\ & s_{\mathrm{B}}=12 t^2-25 t \end{aligned} $$
$$ \begin{aligned} S_{\mathrm{A}} & =S_{\mathrm{B}} \\ 12 t^2-\frac{4}{3} t^3-20 t & =12 t^2-25 t \\ -\frac{4}{3} t^3+5 t & =0 \\ -t\left(\frac{4}{3} t^2-5\right) & =0 \\ \frac{4}{3} t^2 & =5 \\ t & =\sqrt{\frac{15}{4}} \\ & =1.936 \mathrm{~s} \end{aligned} $$
(c) $$ \begin{aligned} & v_{\mathrm{A}}=0 \\ & 24 t-4 t^2-20=0 \\ & t^2-6 t+5=0 \\ & (t-1)(t-5)=0 \\ & t=1 \mathrm{~s}, 5 \mathrm{~s} \end{aligned} $$
$$ \begin{aligned} s & =\left|\int_0^1\left(24 t-4 t^2-20\right) \mathrm{d} t\right|+\int_1^5\left(24 t-4 t^2-20\right) \mathrm{d} t \\ & =\left|\left[12 t^2-\frac{4}{3} t^3-20 t\right]_0^1\right|+\left[12 t^2-\frac{4}{3} t^3-20 t\right]_1^5 \\ & =\left|\left[12(1)^2-\frac{4}{3}(1)^3-20(1)\right]-\left[12(0)^2-\frac{4}{3}(0)^3-20(0)\right]\right|+ \\ & \quad\left[\left[12(5)^2-\frac{4}{3}(5)^3-20(5)\right]-\left[12(1)^2-\frac{4}{3}(1)^3-20(1)\right]\right] \\ & =\frac{28}{3}+\frac{128}{3} \\ & =52 \mathrm{~m} \end{aligned} $$