Post by: hungover on March 27, 2013, 09:15:41 pm
how do you do it? can someone please solve it and explain.
The question is attached below.
Post by: Lasercookie on March 27, 2013, 10:59:35 pm
This question hints directly at that method too.
If they're travelling the same speed, that's when they intersect. Find the distance? Well that's the area under the v/t curve.
When does the policeman overtake, that's when the position of the cars are the same.
So a) Easy, draw out the constant velocity of the speeding car, and then draw out what the motorcycles velocity would be (so that would start at t = 10 and go up linearly until it reaches v = 40m/s after which it starts driving at a constant velocity.
b) At v = 30m/s they're travelling at the same speed. This isn't necessarily when the cars are at the same position. You can see this visually. At what point? I'm not too sure what this question wants of us. Well you could describe it in terms of time, this is something useful to find for later questions so we might as well do it. It's 10 seconds + the time it took for the motorcycle to reach 30 m/s from rest. So we can use v = u + at and see that it would be
Note how these will form the lengths of the shapes you graphed. You could also describe it in terms of displacement, relative to some starting point.
c) What is the maximum distance the speeding car is ahead of motorcycle. Well once the motorcycle starts moving at t = 10, they begin to get closer and closer together, since the motorcycle is accelerating and the speeding car is not. You might want to double check that reasoning just to double check, because I have a nagging feeling that I might be oversimplifying this. So this is probably just asking for the displacement of the car from t = 0 to t = 10. Note that this will be the area of a square.
d) So they overtake when the positions are the same, that is the area under their velocity-time curves are the same (since the integral of velocity is displacement). Let the time when they overtake be equal to t, and you can use that variable to figure out the lengths of the shapes.
e) So this is basically asking how far the speeding car travel from t = 0 to t = the time you just figured out in part d)
You can also tackle these algebraically, but you might find that doing it graphically makes it a lot clearer. You might note how for these simpler shapes at least where you don't really need to integrate and can just do it with simple area formulas instead etc. how the area of square might match up to terms such as x = vt etc.
Post by: hungover on March 28, 2013, 06:01:40 pm