Question

If you walk 1.4 km north and then 4.8 km east, what are the magnitude and direction of your resultant displacement?

Answer 1

c^2 = a^2 + b^2 = 1.4^2 + 4.8^2 = 1.96 + 23.04 = 25

c = √25 = 5

sin of angle = opp/hyp = 4.8 / 5 = 0.96
sin ^-1 (0.96) = 73.7398 ...

magnitude: 5 km; direction: 73.7° east of north

Answer 2

The magnitude and direction of the resultant displacement is 5 km and 16.26° North -east.

Step-by-step explanation:

Given that,

Distance in north = 1.4 km

Distance in east = 4.8 km

We need to calculate the magnitude of resultant displacement

Using formula of displacement

`(AC)^2=AB^2+BC^2`

`AC=√((AB)^2+(BC)^2)`

`AC=√((1.4)^2+(4.8)^2)`

`AC=5\ km`

We need to calculate the direction

Using formula of direction

`\tan\theta=(y)/(x)`

`\tan\theta=(1.4)/(4.8)`

`\theta=\tan^(-1)0.29167`

`\theta=16.26^(\circ)\ N-E`

Hence, The magnitude and direction of the resultant displacement is 5 km and 16.26° North -east.