A car is driven east for a distance of 42 km, then north for 25 km, and then in

Question

A car is driven east for a distance of 42 km, then north for 25 km, and then in a direction 34° east of north for 25 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction. A car is driven east for a distance of 42 km, then north for 25 km, and then in a direction 34° east of north for 25 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction. A car is driven east for a distance of 42 km, then north for 25 km, and then in a direction 34° east of north for 25 km. Determine (a) the magnitude of the car's total displacement from its starting point and (b) the angle (from east) of the car's total displacement measured from its starting direction.

Explanation / Answer

We begin by writing the three vectors in component form and proceed to add them and compute the sum in component, magnitude and direction form.

a = 42km ˆ i + 0 ˆ j

b = 0 ˆ i + 25 km ˆ j

c = 25sin 34°km ˆ i + 25 cos 34°km ˆ

R = (42 + 25sin 34°)km ˆ i + (25 + 25cos 34°)jkm ˆ

=55.97i+45.72j

R=sqrt(55.97^2+45.72^2)

R=72.26km

tan thetha=45.72/55.97

=39.91 degrees

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