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)`

Step-by-step 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|=√(\left ( 65sin78\right )^2+\left ( 42+65cos78\right )^2)`

`|r_2|=√(63.58^2+55.514^2)`

`|r_2|=84.405 m`

for direction

`tan\theta =(42+65cos78)/(65sin78)`

`tan\theta =0.8731`

`\theta =41.124^(\circ)with\ respect\ to\ east`

`\theta =48.876^(\circ)with\ respect\ to\ North`