A boy walks uniformly along the sides of a re | vedclass

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Question

(B) The boy moves from corner $A$ to $B$ and then to $C$. Distance travelled is the actual path length: $Distance = AB + BC = 400 \, m + 300 \, m = 70

Vedclass · Accepted Answer

His displacement is $700 \, m$.

Vedclass · Answer

(B) The boy moves from corner $A$ to $B$ and then to $C$. Distance travelled is the actual path length: $Distance = AB + BC = 400 \, m + 300 \, m = 700 \, m$. Displacement is the shortest distance between the initial point $A$ and final point $C$: $\vec{AC} = \vec{AB} + \vec{BC}$ $|\vec{AC}| = \sqrt{(AB)^2 + (BC)^2} = \sqrt{(400)^2 + (300)^2} = \sqrt{160000 + 90000} = \sqrt{250000} = 500 \, m$. Since the boy changes direction at corner $B$,his velocity vector changes,meaning his velocity is not uniform. Therefore,the statement 'His displacement is $700 \, m$' is incorrect.

$A$ boy walks uniformly along the sides of a rectangular park of size $400 \, m \times 300 \, m$,starting from one corner to the diagonally opposite corner. Which of the following statements is incorrect?

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The position vector of a particle is given by the expression $\vec{r} = 3t^2 \hat{i} + 4t^2 \hat{j} + 7 \hat{k}$. The distance traversed in the first $10 \, s$ is ......... $m$.

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