Day: February 9, 2023

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202302091224 Exercise 20.2

(All accelerations are to be taken as uniform and in a straight line.)

1. A particle starts with velocity 3\,\mathrm{m\,s^{-1}} and accelerates at 0.5\,\mathrm{m\,s^{-2}}. What is its velocity after i. 3\,\mathrm{s}, ii. 10\,\mathrm{s}, iii. t\,\mathrm{s}? How far has it travelled in these times?
2. A body, decelerating at 0.8\,\mathrm{m\,s^{-2}}, passes a certain point with a speed of 30\,\mathrm{m\,s^{-1}}. Find its velocity after 10\,\mathrm{s}, the distance covered in that time and how much further the body will go until it stops.
3. A particle travelling with acceleration of 0.75\,\mathrm{m\,s^{-2}} passes a point O with speed 5\,\mathrm{m\,s^{-1}}. How long will it take to cover a distance of 250\,\mathrm{m}? What will its speed be at that time?
4. If a particle passes a certain point with speed 5\,\mathrm{m/s} and is accelerating at 3\,\mathrm{m/s^{2}} how far will it travel in the next 2\,\mathrm{s}? How long will it take (from the start) to travel 44\,\mathrm{m}?
5. A body falls from rest with an acceleration of 10\,\mathrm{m\,s^{-2}}. What is its velocity 5\,\mathrm{s} afterwards? How far has it fallen by then?

Extracted from A. Godman & J. F. Talbert. (1975). Additional Mathematics Pure and Applied in SI Units.

Roughwork.

1.

\begin{aligned} v(t) & = u+at = 3 + 0.5t \\ v(3) & = 3+0.5(3) = 4.5\,\mathrm{m\,s^{-1}}\\ v(10) & = 3+0.5(10) = 8\,\mathrm{m\,s^{-1}} \\ \end{aligned}

\begin{aligned} s(t) & =ut+\frac{1}{2}at^2 = 3t+\frac{t^2}{4} \\ s(3) & = 3(3)+\frac{(3)^2}{4} = 11.25\,\mathrm{m}\\ s(10) & = 3(10)+\frac{(10)^2}{4} = 55\,\mathrm{m}\\ \end{aligned}

2.

Not to be attempted.

3.

\begin{aligned} s & = ut+\frac{1}{2}at^2 \\ (250) & = (5)t+\frac{1}{2}(0.75)t^2 \\ 0 & = 3t^2 + 40t - 2000 \\ 0 & = (3t+100)(t-20) \\ t & = 20\,\mathrm{s}\textrm{ \scriptsize{OR} }-\frac{100}{3}\,\mathrm{s}\textrm{ (rej.)} \\ \end{aligned}

\begin{aligned} v & = u+at \\ & = (5)+(0.75)(20) \\ & = 20\,\mathrm{m\,s^{-1}} \\ \end{aligned}

or,

\begin{aligned} v^2 & =u^2+2as \\ v & = \pm\sqrt{(5)^2+2(0.75)(250)} \\ & = 20\,\mathrm{m\,s^{-1}}\textrm{ \scriptsize{OR} }-20\,\mathrm{m\,s^{-1}}\textrm{ (rej.)} \\ \end{aligned}

4. Not to be attempted.

5. Not to be attempted.