A $2 \mathrm{~m}$ wide truck is moving with a uniform speed $v_{0}=8 \mathrm{~m} / \mathrm{s}$ along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed $v$ when the truck is $4 \mathrm{~m}$ away from him. The minimum value of $v$ so that he can cross…

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A $2 \mathrm{~m}$ wide truck is moving with a uniform speed $v_{0}=8 \mathrm{~m} / \mathrm{s}$ along a straight horizontal road. A pedestrian starts to cross the road with a uniform speed $v$ when the truck is $4 \mathrm{~m}$ away from him. The minimum value of $v$ so that he can cross the road safely is

Aicharya · Accepted Answer

A truck which is $2 m$ wide is moving with a uniform velocity along a straight line. At the same time a pedestrian starts to cross the road. Let us assume that person is in the position $\mathrm{C}$, he wants to cross the road and reach point $B$. the distance between the truck and the person is $4 m$. Let $A$ be the point parallel to the person at $C$ at a distance $2 m$. With this draw a triangle joining the points $A, B$ and $C$. Then $A C$ is $2 m$, let $A B$ be $x$ and let $B C$ be y. Let us make a diagram using above discussion, The distance between the truck and the point $\mathrm{B}, s=4+x$ The truck moves with the velocity, $v_{T}=8 m s^{-1}$ Let us take the triangle $\mathrm{ABC}$, it is a right angled triangle. According to the Pythagoras theorem, $\begin{array}{l} \Rightarrow y=\sqrt{x^{2}+2^{2}} \ y=\sqrt{x^{2}+4} \end{array}$ The distance between the person at the point $\mathrm{C}$ and the point $\mathrm{B}$ (other side), $y=\sqrt{x^{2}+4}$ The person moves with the velocity $v_{P}=?$ Velocity: the velocity is determined by the rate of change of displacement. I.e. displacement by time taken. $\Rightarrow v=\frac{d}{t}$ This can be rewritten as $\Rightarrow t=\frac{d}{v}$ $V$ is the velocity d is the displacement $t$ is the time taken The time taken by the truck to reach the point $B$ is given by $\begin{array}{l} \Rightarrow t=\frac{d}{v} \ \Rightarrow t_{T}=\frac{4+x}{8} \rightarrow 1 \end{array}$ The time taken by the person to reach the point $B$ is given by $\begin{array}{l} \Rightarrow t=\frac{d}{v} \ \Rightarrow t_{P}=\frac{\sqrt{x^{2}+4}}{v_{p}} \rightarrow 2 \end{array}$ To cross the road safely without getting hit by the truck the time taken by the truck to reach the point B should be equal to the time taken by the person to reach the point B So, equating 1 and 2 $\begin{array}{l} \Rightarrow \frac{4+x}{8}=\frac{\sqrt{x^{2}+4}}{v_{p}} \ \Rightarrow v_{p}=\frac{\sqrt{x^{2}+4}}{4+x} \times 8 \rightarrow 3 \end{array}$ To get minimum velocity of the person $\begin{array}{l} \Rightarrow \frac{d v}{d t}=0 \ \Rightarrow \frac{d}{d t} \frac{\sqrt{x^{2}+4}}{4+x} \times 8=0 \end{array}$ $\Rightarrow 8\left(\frac{(4+x) \times \frac{x}{\sqrt{x^{2}+4}}-\sqrt{x^{2}+4}}{(x+4)^{2}}\right)=0$ $\Rightarrow 8\left(\frac{(4+x) \times \frac{x-\left(x^{2}+4\right)}{\sqrt{x^{2}+4}}}{(x+4)^{2}}\right)=0$ $\Rightarrow 8\left(\frac{(x+4) x-\left(x^{2}+4\right)}{\sqrt{x^{2}+4}}\right)=0$ $\begin{array}{l} \Rightarrow(x+4) x-\left(x^{2}+4\right)=0 \ \Rightarrow x^{2}+4 x-x^{2}-4=0 \ \Rightarrow 4 x-4=0 \ \Rightarrow 4(x-1)=0 \ \Rightarrow(x-1)=0 \ \Rightarrow x=1 \end{array}$ Substitute the value of $x$ in equation 3 $\begin{array}{l} \Rightarrow v_{p}=\frac{\sqrt{x^{2}+4}}{4+x} \times 8 \ \Rightarrow v_{p}=\frac{\sqrt{1^{2}+4}}{4+1} \times 8 \ \Rightarrow v_{p}=\frac{\sqrt{5}}{5} \times 8=\frac{8}{\sqrt{5}} \ \Rightarrow v_{p}=3.57 \mathrm{~ms}^{-1} \end{array}$ The minimum value of velocity required to cross the road safely is $3.57 \mathrm{~ms}^{-1}$ ^

A \(2 \mathrm{~m}\) wide truck is moving with a uniform speed \(v_{0}=8 \mathrm{~m} / \mathrm{s}\) along a…

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