Wandering home I was crossing the road, and a car approached reasonably fast. So I wrote a quick mechanics question and fired it out to my Year 12s (during the holidays!)
Yes, I’m e-mailing you over the holidays. Yes, it’s because I had a geeky moment. So basically, I was crossing the road. A car was approaching reasonably quickly, so I thought: what angle could I cross the road at and not get hit?
I obviously didn’t try it in practice, but it is a good mechanics/vectors question 🙂
I’ve attached a photo of the question. Have a go! Any correct answers will win merits and stickers. There may be other chocolate-based prizes on offer.
Hope you’re all having a great holiday,
<I’ve lost the photo of the question!>
The first response, after an hour, was this:
I quite liked this mechanics question as it is simple to understand whilst still maintaining its difficulty.
i) You’ll want to consider how long it takes to cross the road, compared to how long it will take the car to reach you. First off all it will take:
7 / 2.5cos20 = 2.98s to cross the road.
The relative velocity of the car towards you is 14 – 2.5sin20. We want to find the time the car takes to catch up 40m.
t = 40/ (14-2.5sin20) = 3.04s. This is greater than the time taken to cross the road, so you are good to go!
ii) More generally we are looking to solve the inequality:
40/(14-2.5sinx) > 7/(2.5cosx) for acute x (or x between -90 and 0 degrees)
In the whole range of possible values of x (14-2.5sinx) and 2.5cosx are positive. So this leads to:
200cosx + 35sinx > 196, which by the R-alpha method is the same as:
203cos(x – arctan(7/40)) > 196
cos(x – 9.93 degrees) > 0.965
You might want to imagine the cosine graph now, meaning this leads to (d means degrees):
x – 9.93d < 15.1d or x-9.93d > -15.1d
And we find:
-5.21 degrees < x < 25.1 degrees.
So there you go Mr T… next time you find yourself crossing the road, cross towards the oncoming traffic.
Thought I would share it with you all. Love a good response to a question!