Question

A car is driven east for a distance of 47 km, then north for 28 km, and then in a direction 35° east of north for 27 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.

Answer 1

(a) 82 Km

(b) 32°

Step-by-step explanation:

First you should draw the vectors in the cartesian plane (please see the picture below).

As the car is driven east for a distance of 47 Km, your first vector should be drawn from the origin (0,0) and on the x axis

Then the car is driven north for a distance of 28 Km, so your second vector should be drawn from the origin and on the y axis.

And the car finally goes east of north for 27Km, so the third vector should be drawn from the origin east of north forming an 35° angle with x axis.

Then you should find the components of the vector in x and y:

For Vector 1 (
`V_(1)`)

`V_(1x)=47Km`

`V_(1y)=0`

For Vector 2 (
`V_(2)`)

`V_(2x)=0`

`V_(2y)=28Km`

For Vector 3 (
`V_(3)`)

`V_(3x)=27cos37^(o)`

`V_(3x)=21.56`

`V_(3y)=27sin37^(o)`

`V_(3y)=16.25`

To find the magnitud of the car´s total displacement, (R) you should add up all the x and y components.

For the x component:

`R_(x)=V_(1x)+V_(2x)+V_(3x)`

`R_(x)=47+0+21.56`

`R_(x)=68.56Km`

For the x component:

`R_(y)=V_(1y)+V_(2y)+V_(3y)`

`R_(y)=0+28+16.25`

`R_(y)=44.25Km`

Now please see the second picture that is showing the components x and y as the sides of a right triangle, and we are going to use the Pythagorean theorem to find the resultant, R.

`R=\sqrt{R_(x)^(2)+R_(y)^(2)`

`R=\sqrt{68.56^(2)+44.25^(2)}`

`R=82Km`

And to find the angle of the car´s total displacement (α), we use the same right triangle with the relationship between its legs.

`tan(\alpha)=(44.25)/(68.56)`

`tan(\alpha)=0.64`

`\alpha=tan^(-1)(0.64)`

`\alpha =32^(o)`