Question

Walk 42 miles due north, deviate 78 degrees to east, and walk 65miles. What is the displacement? ( magnitude and direction with respect to North). A) Show work through calculations for predictions

Answer 1

Answer:84.405m,$\theta =48.876^{\circ}$

Explanation:

Given

Person walk 42 miles due to north  so its position vector is

$r_1=42\hat{j}$

Now he deviates $78^{\circ}$ to east and walk 65 miles

so its new position vector

$r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}$

$r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}$

So magnitude of acceleration is

$|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}$

$|r_2|=\sqrt{63.58^2+55.514^2}$

$|r_2|=84.405 m$

for direction

$tan\theta =\frac{42+65cos78}{65sin78}$

$tan\theta =0.8731$

$\theta =41.124^{\circ}with\ respect\ to\ east$

$\theta =48.876^{\circ}with\ respect\ to\ North$