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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A hiker begins a trip by first walking 25 km southeast from her car. She stops and sets up her tent for the night. On the second day, she walks 40 km in a direction 60$^{\circ}$ north of east, at which point she discovers a forest ranger’s tower. Write the total displacement in unit vector form.

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Answer: R = 37.7 $\hat {i}$ + 16.9 $\hat {j}$
We need to determine the components of the hiker’s displacement for each day.
We can denote the displacement on day 1 as A and day 2 as B.
On Day 1, Displacement A has a magnitude of 25 km and is directed 45$^{\circ}$ below the positive x axis.
$\Rightarrow A_x = A \cos (-\theta) = 25 \times \cos (-45^{\circ}) = 25 \times 0.707 = 17.7\; km$
$\Rightarrow A_y = A \sin (-\theta) = 25 \times \sin (-45^{\circ}) = 25 \times - 0.707 = -17.7\; km$
On Day 2, Displacement B has a magnitude of 40 km and is directed 60$^{\circ}$ above the positive x axis.
$\Rightarrow B_x = B \cos (\theta) = 40 \times \cos (60^{\circ}) = 40 \times 0.5 = 20\; km$
$\Rightarrow B_y = B \sin (\theta) = 40 \times \sin (60^{\circ}) = 40 \times 0.866 = 34.6\; km$
Now, $R_x = A_x + B_x = 17.7 + 20 = 37.7$ and $R_y = A_y + B_y = -17.7 + 34.6 = 16.9$
$\Rightarrow R = 37.7 \hat {i} + 16.9 \hat {j}$
answered Aug 11, 2014 by balaji.thirumalai
edited Aug 11, 2014 by balaji.thirumalai
 

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