Answer:84.405m,[tex]\theta =48.876^{\circ}[/tex]
Explanation:
Given
Person walk 42 miles due to north so its position vector is
[tex]r_1=42\hat{j}[/tex]
Now he deviates [tex]78^{\circ}[/tex] to east and walk 65 miles
so its new position vector
[tex]r_2=42\hat{j}+65cos78\hat{j}+65sin78\hat{i}[/tex]
[tex]r_2=65sin78\hat{i}+\left ( 42+65cos78\right )\hat{j}[/tex]
So magnitude of acceleration is
[tex]|r_2|=\sqrt{\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2}[/tex]
[tex]|r_2|=\sqrt{63.58^2+55.514^2}[/tex]
[tex]|r_2|=84.405 m[/tex]
for direction
[tex]tan\theta =\frac{42+65cos78}{65sin78}[/tex]
[tex]tan\theta =0.8731 [/tex]
[tex]\theta =41.124^{\circ}with\ respect\ to\ east[/tex]
[tex]\theta =48.876^{\circ}with\ respect\ to\ North[/tex]
