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A tractor travels in a straight line from a point $A$ to a point $B$. The velocity of the tractor is $vm{\text{ }}{s^{ - 1}}$ at time $t\,s$ after leaving $A$.

a) The diagram shows an approximate velocity-time graph for the motion of the tractor. The graph consists of two straight line segments. Use the graph to find an approximation for

(i) the distance $AB$,

(ii) the acceleration of the tractor for ${\text{0 \lt t \lt 400}}$ and for ${\text{400 \lt t \lt 800}}$.

b) The actual velocity of the tractor is given by $v = 0.04t - 0.00005{t^2}$ for $0 \leqslant t \leqslant 800$.

(i) Find the values of $t$ for which the actual acceleration of the tractor is given correctly by the approximate velocity-time graph in part (a).

For the interval $0 \leqslant t \leqslant 400$, the approximate velocity of the tractor in part (a) is denoted by ${v_1}{\text{ }}m{\text{ }}{s^{ - 1}}$.

(ii) Express ${v_1}$ in terms of $t$ and hence show that ${v_1} - v = 0.00005{\left( {t - 200} \right)^2} - 1$.

(iii) Deduce that $ - 1 \leqslant {v_1} - v \leqslant 1$.

پاسخ تشریحی :
نمایش پاسخ

a)(i) $\left[ {2 \times {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 2$}}\left( {1 + 9} \right)400} \right]$

Approximation is $4000{\text{ }}m$

(ii) Accelerations are $0.02{\text{ }}m{s^{ - 2}}$ and $ - 0.02{\text{ }}m{s^{ - 2}}$

b)(i) $0.04 - 0.0001t =  \pm 0.02$

Values of t are 200 and 600

(ii) ${v_1} - v = 0.02t + 1 - 0.04t + 0.00005{t^2}$

${v_1} - v = \left[ {0.00005{t^2} - 0.02t + 2 - 1} \right]$

$ = 0.00005\left( {{t^2} - 400t + 40000} \right) - 1$

$ = 0.00005{\left( {t - 200} \right)^2} - 1$

(iii) For using ${\left( {{v_1} - v} \right)_{min}}$ occurs when

$t = 200 \to  - 1 \leqslant {v_1} - v$

For using ${\left( {{v_1} - v} \right)_{\max }}$ occurs when $t = 0$ and

when $t = 400 \to {v_1} - v \leqslant 1$

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