Question

A man walks 1.80 km north and then 2.75 km east, all in 3.00 hours.(a)What is the magnitude (in km) and direction (in degrees north of east) of his displacement during the given time?magnitude km direction ° north of east(b)What is the magnitude (in km/h) and direction (in degrees north of east) of his average velocity during the given time?magnitude km/h direction ° north of east(c)What was his average speed (in km/h) during the same time interval?km/h

Answer 1

1. Displacement = 3.29 km

b. His direction is
`33.21^(o)` north of east

2. Average velocity is 1.10 km/h

b. His direction is
`33.21^(o)` north of east

3. Average speed = 1.52 km/h

Step-by-step explanation:

1. Displacement =
`\sqrt{x^(2) + y^(2) }`

=
`\sqrt{1.8^(2) + 2.75^(2) }`

=
`√(10.8025)`

= 3.2867

Displacement = 3.29 km

b. tan θ =
`(y)/(x)`

θ =
`tan^(-1) (1.8)/(2.75)`

=
`tan^(-1)`0.6546

= 33.20878

θ =
`33.21^(o)`

His direction is
`33.21^(o)` north of east

2. Average velocity =
`(displacement)/(time)`

=
`(3.29)/(3)`

= 1.09667

Average velocity is 1.10 km/h

b. His direction is
`33.21^(o)` north of east.

3. average speed =
`(total distance)/(time)`

total distance covered = 1.80 km + 2.75 km

= 4.55 km

Average speed =
`(4.55)/(3)`

= 1.5167

Average speed = 1.52 km/h