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A walker travels along a straight road passing through the points $A$ and $B$ on the road with speeds $0.9{\text{ }}m{\text{ }}{s^{ - 1}}$ and $1.3{\text{ }}m{\text{ }}{s^{ - 1}}$ respectively. The walker’s acceleration between $A$ and $B$ is constant and equal to $0.004{\text{ }}m{\text{ }}{s^{ - 2}}$.

a) Find the time taken by the walker to travel from $A$ to $B$, and find the distance $AB$.

A cyclist leaves $A$ at the same instant as the walker. She starts from rest and travels along the straight road, passing through $B$ at the same instant as the walker. At time ts after leaving $A$ the cyclist’s speed is $k{t^3}{\text{ }}m{\text{ }}{s^{ - 1}}$, where $k$ is a constant.

b) Show that when $t = 64.05$ the speed of the walker and the speed of the cyclist are the same, correct to 3 significant figures.

c) Find the cyclist’s acceleration at the instant she passes through $B$.

پاسخ تشریحی :
نمایش پاسخ

a) $[1.3 = 0.9 + 0.004T,$

${1.3^2} = {0.9^2} + 2 \times 0.004S]$

Time is $100{\text{ }}s$ (or distance is $110{\text{ }}m$)

Distance is $110{\text{ }}m$ (or time is $100{\text{ }}s$)

b) $\int {k{t^3}dt}  = {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}k{t^4}$

$\left[ {k\left( {{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle 4$}}{{100}^4} - 0} \right) = 110} \right]$

$k = 4.4 \times {10^{ - 6}}$

$[{v_W} = 0.9 + 0.004 \times 64.05,$

${v_C} = 4.4 \times {10^{ - 6}} \times {64.05^3}]$

Both are equal to $1.16{\text{ }}m{s^{ - 1}}$ correct to 3 sf.

c) Acceleration $ = 3k{t^2}$

Acceleration at $B$ is $0.132{\text{ }}m{s^{ - 2}}$

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