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Grade 11Mechanics

a 2m wide truck is moving with a uniform speed v0=8m/s along a straight horizontal road a pedestrian starts to cross the road with a uniform speed v when the truck is 4m away from him the minimum value of v so that he can cross the road safely is

Profile image of younus jalil
11 Years agoGrade 11
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2 Answers

Profile image of Naveen Kumar
11 Years ago

Please see the image attached:
260-1786_pedestrain.png
Profile image of Simran Banga
9 Years ago
Let the man crossing the road at an angle θθ. For safe crossing the man must cross the road by the time the truck describes a distance 4+AC4+ACor 4+2cotθ4+2cot⁡θ=>AC+4v0=BCv=>AC+4v0=BCv=4+2cotθ8=2/sinθv=4+2cot⁡θ8=2/sin⁡θvv=82sinθ+cosθv=82sin⁡θ+cos⁡θfor minimum timedvdθdvdθ=0=>−8(2cosθ−sinθ)(2sinθ+cosθ)2=0=>−8(2cos⁡θ−sin⁡θ)(2sin⁡θ+cos⁡θ)2=0=02cosθ−sinθ=02cos⁡θ−sin⁡θ=0tanθ=2tan⁡θ=2=>sinθ=25–√=>sin⁡θ=25=>cosθ=15–√=>cos⁡θ=15Therefore v=82×25√−15√v=82×25−15=85–√

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