Kinematics Question 659
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 it’s 4 m away from him. The minimum value of v so that he can cross the road safely it’s
Options:
A) 2.62 m/s
B)4.6 m/s
C) 3.57 m/s
D)1.414 m/s
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Let the man starts crossing the road at an angle $ \theta $ .
For safe crossing the condition it’s 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} $
