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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A car travels 20 km due north and then 35 km in a direction 60$^{\circ}$ west of north as shown below. Find the magnitude and direction of the car’s resultant displacement.

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Magnitude = 48.2 km $\quad$ Direction = 39$^{\circ}$
To calculate magnitude R, we can use the law of cosines as applied to the triangle.
The internal angle $\theta= $180$^{\circ}$ - 60$^{\circ}$ = 120$^{\circ}$
We know that $R^2 = A^2 + B^2 - 2AB\cos \theta$
$\Rightarrow R = \sqrt { A^2 + B^2 - 2AB \cos \theta}$
$\Rightarrow R = \sqrt { 20^2 + 35^2 - 2\times 20 \times 35 \times \cos 120^{\circ}} = 48.2\;$km
The direction of R measured from the northerly direction can be obtained from the law of sines:
$\large\frac{\sin x}{B} $$ = \large\frac{\sin \theta}{R}$
$\Rightarrow x = \sin^{-1} \big ( 35 \times \large\frac{\sin 120^{\circ}} {48.2} \big )$
$\quad = \sin^{-1} (0.629) = 39^{\circ}$
answered Aug 11, 2014 by balaji.thirumalai

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