Question

Consider two displacements, one of magnitude 15 m and another of magnitude 20 m. What angle between the directions of this two displacements give a resultant displacement of magnitude (a) 35 m, (b) 5 m, and (c) 25 m.

Answer 1

Answer:

a) 0°

b) 180°

c) 90°

Explanation:

Hello!

To solve this question let a be the vector whose length is 15 m and b the vector of length 20 m

So:

|a | = 15

|b | = 20

Since we are looking for the angle between the vectors we need to calculate the length of the sum of the two vectors, this is:

$|a+b|^{2} = |a|^{2} + |b|^{2} + 2 |a||b|cos(\theta)$

Now we replace the value of the lengths:

$|a+b|^{2} = 15^{2} + 20^{2} + 2*15*20*cos(\theta)$

$|a+b|^{2} = 625 + 600*cos(\theta)$ --- (1)

Now, if:

a) |a+b| = 35

First we can see that 20 + 15 = 35, so the angle must be 0, lets check this:

$35^{2} = 625 + 600*cos(\theta)$

$1225 = 625 + 600*cos(\theta)$

$600 = 600*cos(\theta)$

$1= cos(\theta)$

and :

$\theta = arccos(1)$

      θ = 0

b) |a+b|=5

From eq 1 we got:

$\theta = arccos(\frac{|a+b|^{2}-625}{600})$   --- (2)

$\theta = arccos(\frac{|a+b|^{2}-625}{600})$

$\theta = arccos(-1)$

  θ = π     or    θ = 180°

c) |a+b|=25

$\theta = arccos(\frac{|25|^{2}-625}{600})$

$\theta = arccos(-1)$

  θ = π/2     or    θ = 90°