Sec 4 A Math: Kinematics, practice questions & worked solutions

(a)

$$\begin{aligned} 12t-\frac{3}{2}t^2 &= 0\\ t\!\left(12-\frac{3}{2}t\right) &= 0\\ t=0 \;&\text{or}\; 8\\ \therefore\ &8\ \text{s} \end{aligned}$$

(b)

$$\begin{aligned} \text{distance} &= \int_0^8 12t-\frac{3}{2}t^2\,\mathrm{d}x\\ &= \left[6t^2-\frac{1}{2}t^3\right]_0^8\\ &= [128]-[0]\\ &= 128\ \text{m} \end{aligned}$$

(c)

$$\begin{aligned} a &= 12-3t\\ \text{When } t=5,\ a &= -3\ \text{m/s}^2 \end{aligned}$$

(a)

$$\begin{aligned} v &= k\cos 2t\\ a = \frac{dv}{dt} &= -2k\sin 2t\\ t = \frac{\pi}{12}, \ a &= -6\\ -6 &= -2k\sin 2\left(\frac{\pi}{12}\right) \end{aligned}$$ When $$\begin{aligned} -3 &= -k\sin\left(\frac{\pi}{6}\right)\\ 3 &= k\left(\frac{1}{2}\right)\\ k &= 6 \end{aligned}$$

(b)

At rest $\Rightarrow v = 0$ $$\begin{aligned} v &= 6\cos 2t\\ 6\cos 2t &= 0\\ \cos 2t &= 0\\ 2t &= \frac{\pi}{2}\\ t &= \frac{\pi}{4} \end{aligned}$$

(c)

$$\begin{aligned} \text{Distance from } O \text{ to } P &= \int_0^{\frac{\pi}{4}} 6\cos 2t\ dt\\ &= \left[\frac{6\sin 2t}{2}\right]_0^{\frac{\pi}{4}}\\ &= 3\left(\sin\frac{\pi}{2} - \sin 0\right)\\ &= 3\text{ m} \end{aligned}$$

Alternative

$$\begin{aligned} s &= \int 6\cos 2t\ dt\\ &= \frac{6\sin 2t}{2} + c \end{aligned}$$ At $t = 0$, $s = 0$ $$\begin{aligned} \frac{6\sin 2(0)}{2} + c &= 0\\ c &= 0 \end{aligned}$$ At $t = \dfrac{\pi}{4}$, $$s = \frac{6\sin 2\left(\frac{\pi}{4}\right)}{2} = 3$$

(a)

$$\begin{aligned} v &= 10t + c \end{aligned}$$ When $t = 0$, $v = 0$; $$\begin{aligned} \therefore c &= 0\\ v &= 10t \end{aligned}$$ When $t = 5$, $v = 50$.

Velocity $= 50$ m/s

(b)

$$\begin{aligned} s &= 5t^2 + c \end{aligned}$$ When $t = 0$, $s = 0$; $$\begin{aligned} \therefore c &= 0\\ s &= 5t^2 \end{aligned}$$ When $t = 5$, $s = 125$.

Distance $= 125$ cm

(c)

$$\begin{aligned} V &= 10T - k\frac{T^{\frac{3}{2}}}{\frac{3}{2}} + c \end{aligned}$$ When $T = 0$, $V = 50$ $$\begin{aligned} \therefore c &= 50\\ V &= 10T - \frac{2kT^{\frac{3}{2}}}{3} + 50\\ 26 &= 40 - \frac{2k(8)}{3} + 50\\ \therefore k &= 12 \end{aligned}$$

(a)

Since $t>0$, $\dfrac{25}{2t+1}>0$ and hence $v>0$. Therefore particle will never change its direction.

As $t$ increases, $\dfrac{25}{2t+1}$ decreases and approaches $0$ / becomes very small.

Therefore particle will eventually travel at a speed close to $10$ m/s.

(b)

$$\begin{aligned} v &= 25(2t+1)^{-1}+10\\ a &= -25(2t+1)^{-2}(2)\\ &= -\frac{50}{(2t+1)^2}\\ a_2 &= -\frac{50}{25} = -2\ \text{m/s}^2 \end{aligned}$$

(c)

$$s = \frac{25\ln(2t+1)}{2}+10t+c$$ When $t=0$, $s=0$, $c=0$. $$\begin{aligned} s &= \frac{25\ln(2t+1)}{2}+10t\\ s_4 &= \frac{25\ln 9}{2}+40 \approx 67.465\\ s_5 &= \frac{25\ln 11}{2}+50 \approx 79.973 \end{aligned}$$ $$\begin{aligned} \text{Distance travelled} &= 79.973-67.465\\ &= 12.5\ \text{m} \end{aligned}$$

(a)

When $t = 0$, $$v = \cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}\text{ m/s}$$

(b)

When $v = 0$, $$\begin{aligned} \cos\!\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) &= 0\\ \frac{\pi}{4}t + \frac{\pi}{6} = \frac{\pi}{2} \ &\text{or}\ \frac{\pi}{4}t + \frac{\pi}{6} = \frac{3\pi}{2}\\ t = \frac{4}{3} \ &\text{or}\ \frac{16}{3} = 1.33 \text{ or } 5.33 \end{aligned}$$ $$\begin{aligned} \cos\!\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) &= 0\\ \frac{\pi}{4}t + \frac{\pi}{6} = \frac{\pi}{2} \ &\text{or}\ \frac{\pi}{4}t + \frac{\pi}{6} = \frac{3\pi}{2}\\ t = \frac{4}{3} \ &\text{or}\ \frac{16}{3} = 1.33 \text{ or } 5.33 \end{aligned}$$

(c)

$$\begin{aligned} s &= \int\cos\!\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) dt\\ s &= \frac{4}{\pi}\sin\!\left(\frac{\pi}{4}t + \frac{\pi}{6}\right) + c \end{aligned}$$ When $t = 0$, $s = \dfrac{1}{\pi}$, $$\begin{aligned} \frac{1}{\pi} &= \frac{4}{\pi}\left(\sin\frac{\pi}{6}\right) + c\\ \frac{1}{\pi} &= \frac{4}{\pi}\left(\frac{1}{2}\right) + c\\ c &= \frac{1}{\pi} - \frac{2}{\pi} = -\frac{1}{\pi}\\ \therefore s &= \frac{4}{\pi}\left[\sin\!\left(\frac{\pi}{4}t + \frac{\pi}{6}\right)\right] - \frac{1}{\pi} \end{aligned}$$ When $t = 0$ $s = \dfrac{1}{\pi}\text{ m} = 0.31830$ m $$\begin{aligned} \text{When } t = \frac{4}{3} s &= \frac{4}{\pi}\left(\sin\frac{\pi}{2}\right) - \frac{1}{\pi} = \frac{3}{\pi}\text{ m or } 0.95492\text{ m}\\ \text{When } t = 4 s &= \frac{4}{\pi}\left(\sin\frac{7\pi}{6}\right) - \frac{1}{\pi} = -\frac{3}{\pi} \text{ or } -0.95492\text{ m} \end{aligned}$$ $$\text{Total distance travelled} = \frac{2}{\pi} + \frac{3}{\pi} + \frac{3}{\pi} = \frac{8}{\pi}\text{ or } 2.55\text{ m}$$

(a)

$$\begin{aligned} a &= 4t-3\\ v &= \int 4t-3\,\mathrm{d}t\\ &= 2t^2-3t+c \end{aligned}$$ At $t=0$, $v=0$, $c=0$. $$\begin{aligned} v &= 2t^2-3t\\ \text{At } t=1,\ v &= 2(1)^2-3(1)=-1\ \text{m/s} \end{aligned}$$

(b)

$$\begin{aligned} v &= 2t^2-3t\\ \text{At } v=0,\ 2t^2-3t &= 0\\ t(2t-3) &= 0\\ t=0 \;&\text{or}\; t=\frac{3}{2}\\ s &= \int 2t^2-3t\,\mathrm{d}t\\ &= \frac{2}{3}t^3-\frac{3}{2}t^2+c \end{aligned}$$ At $t=0$, $s=0$, $c=0$. $$\begin{aligned} s &= \frac{2}{3}t^3-\frac{3}{2}t^2\\ \text{At } t=\frac{3}{2},\ s &= \frac{2}{3}\!\left(\frac{3}{2}\right)^3-\frac{3}{2}\!\left(\frac{3}{2}\right)^2\\ &= -1\frac{1}{8}\ \text{m} \end{aligned}$$ The distance is $1\dfrac{1}{8}$ m.

(c)

$$\begin{aligned} \text{Total distance} &= \left|\int_0^{1.5} 2t^2-3t\,\mathrm{d}t\right|+\int_{1.5}^5 2t^2-3t\,\mathrm{d}t\\ &= \left|\left[\frac{2}{3}t^3-\frac{3}{2}t^2\right]_0^{1.5}\right|+\left[\frac{2}{3}t^3-\frac{3}{2}t^2\right]_{1.5}^5\\ &= \left|\left[\frac{2}{3}(1.5)^3-\frac{3}{2}(1.5)^2-0\right]\right|+\left[\frac{2}{3}(5)^3-\frac{3}{2}(5)^2\right]-\left[\frac{2}{3}(1.5)^3-\frac{3}{2}(1.5)^2\right]\\ &= \left|-\frac{9}{8}\right|+\frac{275}{6}-\left(-\frac{9}{8}\right)\\ &= 48\frac{1}{12} \end{aligned}$$

(a)

$$\begin{aligned} v &= 5 - e^{3 - 2t}\\ 5 - e^{3 - 2t} &= 0\\ e^{3 - 2t} &= 5\\ \ln e^{3 - 2t} &= \ln 5\\ 3 - 2t &= \ln 5\\ t &= 0.695 \end{aligned}$$

(b)

Distance in the first second $$\begin{aligned} &= \left|\int_{0.69528}^{1}\!\!(5 - e^{3-2t})\,dt\right| + \left|\int_0^{0.69528}\!\!(5 - e^{3-2t})\,dt\right|\\[4pt] &= \left|\left[5t - \left(\frac{e^{3-2t}}{-2}\right)\right]_{0.69528}^{1}\right| + \left|\left[5t - \left(\frac{e^{3-2t}}{-2}\right)\right]_0^{0.69528}\right|\\[4pt] &= \left|\left[5(1) + \left(\frac{e^{3-2(1)}}{2}\right)\right] - \left[5(0.69528) + \left(\frac{e^{3-2(0.69528)}}{2}\right)\right]\right|\\[4pt] & + \left|\left[5(0.69528) + \left(\frac{e^{3-2(0.69528)}}{2}\right)\right] - \left[5(0) + \left(\frac{e^{3-2(0)}}{2}\right)\right]\right|\\[4pt] &= 0.3827356 + \left|-4.06636\right|\\ &= 4.449\\ &= 4.45\text{ m} \end{aligned}$$

(c)

$$v = 5 - e^{3 - 2t}$$ As $t$ approaches $\infty$, $e^{3 - 2t}$ approaches $0$. $$v = 5\text{ m/s}$$ The particle’s velocity cannot exceed 5 m/s, and this is the greatest velocity attained.

(a)

$$\begin{aligned} \frac{dv}{dt} &= 6t - 21\\ 6t - 21 &= 0\\ t &= \frac{21}{6}\ \text{s}\\ v &= 3\left(\frac{21}{6}\right)^2 - 21\left(\frac{21}{6}\right) + 18\\ &= -\frac{75}{4}\ \text{ms}^{-1}\\ \frac{d^2v}{dt^2} &= 6 > 0 \end{aligned}$$ Therefore, $v = -\dfrac{75}{4}\ \text{ms}^{-1}$ is the minimum velocity.

(b)

$$\begin{aligned} 3t^2 - 21t + 18 &= 0\\ t^2 - 7t + 6 &= 0\\ (t - 6)(t - 1) &= 0\\ t = 1 \ &\text{or} \ 6 \end{aligned}$$

(c)

$$\begin{aligned} s &= \int 3t^2 - 21t + 18\, dt\\ &= t^3 - \frac{21}{2}t^2 + 18t + c \end{aligned}$$ At $t = 0$, $s = 2$, therefore, $c = 2$ $$s = t^3 - \frac{21}{2}t^2 + 18t + 2$$ At $t = 1$, $$\begin{aligned} s &= 1^3 - \frac{21}{2}1^2 + 18(1) + 2\\ &= 10.5\ \text{m} \end{aligned}$$ At $t = 6$, $$\begin{aligned} s &= 6^3 - \frac{21}{2}(6)^2 + 18(6) + 2\\ &= -52\ \text{m} \end{aligned}$$ At $t = 7$, $$\begin{aligned} s &= 7^3 - \frac{21}{2}(7)^2 + 18(7) + 2\\ &= -43.5\ \text{m} \end{aligned}$$

$$\begin{aligned} \text{Total Distance} &= (10.5 - 2) + 10.5 + 52 + (52 - 43.5)\\ &= 79.5\ \text{m} \end{aligned}$$

(a)

When particle passes $O$, $t = 0$,

Hence, $v = 4 + 8\cos 2(0)$ $$= 12\text{ m/s}$$

(b)

$$\begin{aligned} v &= 4 + 8\cos 2t\\ \frac{dv}{dt} &= -(2)8\sin 2t\\ a &= -16\sin 2t \end{aligned}$$ Least acceleration $$= -16\text{ m/s}^2$$

(c)

$$\begin{aligned} 0 &\leq t \leq 3\\ 0 &\leq 2t \leq 6 \end{aligned}$$ When particle is at rest, $$\begin{aligned} v &= 0\text{ m/s},\\ 4 + 8\cos 2t &= 0\\ \cos 2t &= -\frac{1}{2} \end{aligned}$$ Basic angle $\alpha = \cos^{-1}\left(\dfrac{1}{2}\right) = \dfrac{1}{3}\pi$

For $\cos 2t < 0$, $2t$ lies in Q2, Q3 $$\begin{aligned} 2t &= \pi - \frac{1}{3}\pi,\ \pi + \frac{1}{3}\pi\\ &= \frac{2}{3}\pi,\ \frac{4}{3}\pi\\ t &= \frac{1}{3}\pi,\ \frac{2}{3}\pi \end{aligned}$$

(d)

$$\begin{aligned} s &= \int_0^{\pi/3}(4 + 8\cos 2t)\,dt\\ s &= 4t + 4\sin 2t + c \end{aligned}$$ $$\begin{array}{c|c} t & s\\ \hline 0 & 0\\[2pt] \dfrac{1}{3}\pi & 7.65289\\[6pt] 2 & 4.97279 \end{array}$$ $$\begin{aligned} \text{Distance travelled in the first 2 sec} &= (7.65289 - 0)\\ & + (7.65289 - 4.97279)\\ &= 10.33299\\ &= 10.3\text{ m} \ (\text{to 3 sig.\ fig.}) \end{aligned}$$

Or

$$\begin{aligned} \text{Distance travelled in the first 2 sec} &= \left|\int_0^{\pi/3}(4 + 8\cos 2t)\,dt\right|\\ & + \left|\int_{\pi/3}^{2}(4 + 8\cos 2t)\,dt\right|\\ &= \left|[4t + 4\sin 2t]_0^{\pi/3}\right|\\ & + \left|[4t + 4\sin 2t]_{\pi/3}^{2}\right|\\ &= \left|4\left(\frac{\pi}{3}\right) + 4\left(\sin\frac{2\pi}{3}\right)\right|\\ & + \left|4\left(2 - \frac{\pi}{3}\right) + 4\left(\sin 4 - \sin\frac{2\pi}{3}\right)\right|\\ &= 7.65289 + 2.6801\\ &= 10.33299\\ &= 10.3\text{ m} \ (\text{to 3 sig.\ fig.}) \end{aligned}$$

(a)

When $v=0$, $$\begin{aligned} 2e^{-2t}-e^{-t} &= 0\\ e^{-t}(2e^{-t}-1) &= 0\\ e^{-t}=0\ (\text{NA}) \;&\text{or}\; e^{-t}=\frac{1}{2}\\ t=\ln 2 \;&\text{or}\; 0.693 \end{aligned}$$

(b)

$$\begin{aligned} s &= \int 16(2e^{-2t}-e^{-t})\,dt\\ &= 16(-e^{-2t}+e^{-t})+c \end{aligned}$$ When $t=0$, $s=0$, $c=0$. $$\begin{aligned} s &= 16(-e^{-2t}+e^{-t}) \end{aligned}$$ when $t=\ln 2$, $s=4$ m.

(c)

$$\begin{aligned} a &= 16(-4e^{-2t}+e^{-t}) \end{aligned}$$ when $a=0$, $$\begin{aligned} -4e^{-2t}+e^{-t} &= 0\\ e^{-t}(-4e^{-t}+1) &= 0\\ e^{-t}=0\ (\text{NA}) \;&\text{or}\; 4e^{-t}=1\\ t &= \ln 4 \end{aligned}$$ When $t=\ln 4$, $v=-2$ m/s.