Question

Three displacements through a hedge maze. (b) The displacement vectors. (c) The first displacement vector and its components. (d) The net displacement vector and its components.d1=6.00 mθ1=40°d2=8.00 mθ2=30°d3=5.00 mθ3=0°,where the last segment is parallel to the superimposed x axis. When we reach point c, what are the magnitude and angle of our net displacement →dnet from point i?

Answer 1

18.3 m , 25.4°

Step-by-step explanation:

d1 = 6 m, θ1 = 40°

d2 = 8 m, θ2 = 30°

d3 = 5 m, θ3 = 0°

Write the displacements in the vector form

`d_(1)=6\left ( Cos40\widehat{i}+Sin40\widehat{j} \right )=4.6\widehat{i}+3.86\widehat{j}`

`d_(2)=8\left ( Cos30\widehat{i}+Sin30\widehat{j} \right )=6.93\widehat{i}+4\widehat{j}`

`d_(3)=5\widehat{i}`

The total displacement is given by

`\overrightarrow{d}=\overrightarrow{d_(1)}+\overrightarrow{d_(2)}+\overrightarrow{d_(3)}`

`d=\left ( 4.6+6.93+5 \right )\widehat{i}+\left ( 3.86+4 \right )\widehat{j}`

`d=16.53\widehat{i}+7.86\widehat{j}`

magnitude of resultant displacement is given by

`d ={\sqrt{16.53^(2)+7.86^(2)}}=18.3 m`

d = 18.3 m

Let θ be the angle of resultant displacement with + x axis

`tan\theta =(7.86)/(16.53)=0.4755`

θ = 25.4°