54.7/1009/14/2018 12:10 PM Periodic Tabke Grac 1/2018 10:59 PM Print Calcu lator

Question

54.7/1009/14/2018 12:10 PM Periodic Tabke Grac 1/2018 10:59 PM Print Calcu lator on 15 of 29 An unsuspecting bird is coasting along in an easterly direction at 4.00 mph when a strong wind from the Mapf south imparts a constant acceleration of 0.200 m/s. If the wind's acceleration lasts for 3.50 s, find the magnitude r and direction 8 (measured counterclockwise from the easterly direction) of the bird's displacement over this time interval. (HINT: assume the bird is originally travelling in the +x direction and there are 1609 m in 1 mile.) Number Number Now, assume the same bird is moving along again at 4.00 mph in an easterly direction but this time the acceleration given by the wind is at a 41.0 angle to the original direction of motion. If the magnitude of the acceleration is 0.500 m/s?, find the displacement vector r, and the angle of the displacement, 8 Enter the components of the vector and angle below. (Assume the time interval is still 3.50 s.) Number Number Number O Previous Give Up & View Solution O Check Anwer O Next Exa about us careers privacy policy ch up

Explanation / Answer

u = 4mph=1.78816 m/s

U= 1.78816 i + 0 j

a=0 i + 0.2 j

s= ut + 1/2 at^2

s= (1.78816 i) 3.50 + 0.5 (0.2j) ( 3.5^2)=6.26 i +1.225 j

r= sqroot (6.26^2+ 1.225^2) =3.8 m apprx

angle = tan^-1(1.225/6.26)=11.1 degree from + x axis , north of east

b)U= 1.78816 i + 0 j

a= 0.5 cos 41 i + 0.5 sin 41 j =0.377 i +0.328 j

s= (1.78816 i )3.50 + 0.5 (0.377 i +0.328 j) 3.5^2=6.26 i +2.31 i + 2j =8.57 i + 2 j

r= sqroot ( 8.57^2 + 2^2)= 8.8 m apprx

ang;e = tan^-1 (2/8.57)=13.136 north of east

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