Physics Two Dimensional Motion Question 209
Question: A 2 m wide truck is moving with a uniform speed $ {{\text{v}} _{\text{0}}}\text{= 8 m/s} $ along a straight horizontal road. A pedestrain starts to cross the road with a uniform speed v when the truck is 4 m away from him. The minimum value of v so that he can cross the road safely is
Options:
A) 2.62 m/s B)4.6 m/s C) 3.57 m/s D)1.414 m/s
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Answer:
Correct Answer: C
Solution:
[c] Let the man starts crossing the road at an angle $ \theta $ as shown in figure. For safe crossing the condition is that the man must cross the road by the time the truck describes the distance 4 + AC or $ 4+2cot\theta $ .. $ \
Therefore \frac{4+2\cot \theta }{\text{g}}=\frac{2/\sin \theta }{\text{v}}\text{ or v=}\frac{8}{2\sin \theta +\cos \theta },,…( \text{i} ) $ For minimum v, $ \frac{\text{dv}}{\text{d}\theta }=0 $ $ \text{or},,,\frac{-,8( 2\cos \theta -\sin \theta)}{{{( 2\sin +cos\theta)}^{2}}}=0 $ $ \text{or },,\text{2},\text{cos}\theta -\sin \theta =0\text{ or tan}\theta ,\text{=2} $ $ \text{From equation (i),} $
$ {{\text{v}} _{\min }}=\frac{8}{2( \frac{2}{\sqrt{5}} )+\frac{1}{\sqrt{5}}}=\frac{8}{\sqrt{5}}=3.57\text{ m/s} $
BETA
