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

Answer:

18.3 m , 25.4°

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

$tan\theta =\frac{7.86}{16.53}=0.4755$

θ = 25.4°