HSC motion question (1 Viewer)

[BlueGas]

Here's a HSC question that I attempted and here's my working out for question i), however the success one working out for this question had the answer as Pi, I went further than that using the ASTC rule and was wondering why didn't the success one working out include what I done?

 

[Ekman]

First of all t cannot be 270 as cos(270) = 0, not -1. Also cos is positive in the 1st and 4th quadrant

In terms of your answers, t has to be in radians as they give the boundaries in radians. Hence the answers would be , however we cancel , as it isn't within the boundaries. Thus answer is:

 

[BlueGas]

First of all t cannot be 270 as cos(270) = 0, not -1. Also cos is positive in the 1st and 4th quadrant

In terms of your answers, t has to be in radians as they give the boundaries in radians. Hence the answers would be , however we cancel , as it isn't within the boundaries. Thus answer is:

I get it now, but if cos is positive in the 4th quadrant, you would do 360 -180 which gets you 180. But how do you get 3pi?

 

[InteGrand]

I get it now, but if cos is positive in the 4th quadrant, you would do 360 -180 which gets you 180. But how do you get 3pi?

. But as Ekman said, we discard this solution of as it isn't within the required time range.

(The reason that is that the cosine function has a period of , so if you change its argument by a multiple of , its value is unchanged.)

 

[Ekman]

I get it now, but if cos is positive in the 4th quadrant, you would do 360 -180 which gets you 180. But how do you get 3pi?

Well in trigonometric functions, there are an infinite amount of solutions, for example cos(x)=1 x=0,2pi,4pi,6pi... Hence in order to find solutions that will satisfy t in the question, between 0 to 3pi, we have to factor in not only 1 revolution (from 0 to 2pi) but also another half-revolution (from 2pi to 3pi)