Question

A jogger travels a route that has two parts. The first is a displacement ->A of 2.20 km due south, and the second involves a displacement ->B that points due east.

(a) The resultant displacement ->A + ->B has a magnitude of 3.81 km.
What is the magnitude of B?
______ km
What is the direction of A + B relative to due south?
_____° west of south or east of south?
(b) Suppose that A - B had a magnitude of 3.81 km. What then would be the magnitude of B and what is the direction of A - B relative to due south?

Answer 1

a) B = 3.11 km. θ= 54.7º E of S

b) B = 3.11 km θ= 54.7º W of S

Step-by-step explanation:

a)

• Since we know the value of the total displacement, and the value of the displacement A and its direction, we can find the magnitude of B just applying the Pythagorean Theorem, as follows:

`C=\sqrt{(2.2km)^(2) + B^(2) } = 3.81 Km (1)`

• Solving for B, the only unknown, we get:

`B=\sqrt{(3.81km)^(2) -(2.2km)^(2) } = 3.11 Km (2)`

• Now, applying some simple trig, we can find the angle that (A+B) makes with the S axis, as follows:

`\theta = arc tg ((B)/(A) )= arc tg ( (3.11)/(2.2) )= arctg (1.414) = 54.7 deg (3)`

• Since it's a positive number, applying the convention that the positive angles are measured counterclockwise, this means that this angle is measured East of South.

b)

• If the magnitude of A-B is the same than the one for A+B, this means that the magnitude of B remains the same, i.e. 3.11 Km.
• But if we do graphically A-B, as it is the same as adding A + (-B), we find that the angle of A-B is different to the one in A+B, even the magnitudes of both displacements are the same.
• In this case, B is a negative number, because it's a displacement due west.
• So, applying the same trig that for a) we can find the angle that (A-B) makes with the S Axis, as follows:

`\theta = arc tg ((-B)/(A) )= arc tg ( (-3.11)/(2.2) )= arctg (-1.414) = -54.7 deg (4)`

• So, since it is negative, it's measured clockwise from the S axis, so it's 54.7º W of S.