Question

A particle travels in a straight line with speed v.

The particle slows down and changes direction. The new speed of the particle is v\2

The new velocity has a component of v\4

in the same direction as the initial path of the particle.

Through which angle has the particle turned?

A 27° B 30° C 45° D 60°

Answer 1

D 60°

Step-by-step explanation:

Using trigonometry in the attached image:

`cos\alpha =(v/4)/(v/2)`

`cos\alpha =(1)/(2)`

`\alpha =cos^(-1) (1)/(2)`

Angle=60°

Answer 2

D 60°

Step-by-step explanation:

Using trigonometry:

- The new speed (v/2) of the particle corresponds to the hypothenuse

- The component of v/4 represents the side adjacent to the angle that we want fo find,
`\theta`

So we can write:

`cos \theta = (adjacent)/(hypothenuse)=(v/4)/(v/2)=(1)/(2)`

So we find the angle

`\theta= cos^(-1) ((1)/(2))=60^(\circ)`