Question

please and please answer completion4, completion5, and completion6 also please try to make it clear for my understanding

thank you very much in advance

2.3 Chasing and Colliding Have you ever seen a collision in a computer game? Right silly question. That's most of what happens. Collislons (and avolding them) are critically important in computer graphics, robotics, self-driving cars, automatic lawn mowers, and countless other applications. Suppose we have two robots whose positions are glven by (t and x(). The statement that they collide is expressed mathematically as x(-() at the some time which we might call teollision- (If they have the same position at the same time, then, well.) Example problem Robot 1 starts at an initial position of 100 m and travels with velocity +5 m/s. Robot 2 starts at x = 50 m and travels at +10 m/s, will they collide? If so, when? Solution: First write equations for the positions of each. Here we have constant velocitles so we use We have x,ะเ00 + 5 t and x,-50-101 The collision occurs when X1 = X2 or 100+5t-50+10t Solving for t gives 10 seconds. That's when they collide. But let's check our work. At time t#10 seconds we have x-100+5t-100m+5mls 10s-150m ,-50-10に50m1+10m/s* 10=150m Yes, the positions are indeed the same at t10 seconds. Consider the graph of velocity versus time below. Velocity versus time 09 0.7 06 05 04- 03 02 03 20 12 me 00 COMPLETION 4 First of all, this is not even a function. Explain why it is not a function and fix the problem(s). Hint: What is the velocity at exactly t-2 sec? COMPLETION Your velocity graph is hopefully now a function but it is unphysical. It could not represent real motion and would make computer graphics look obviously fake if used there. Explain why. Hint: describe the motion. COMPLETION 6 Find the displacement of the robot (how far the robot has moved) from time t0 to t10 seconds. xz***2**zx**22

Answer 1

1)The given graph is not a function because in a function, for a certain value of x , there should be a unique value of y but in this case at t= 2 secs, we get multiple values of v which shows that this is not a function.

2) For it to be a function, it should be a discontinuous function such that at t=2 secs, it should have only one value of the velocity let us say 9m/s and this is not possible for the particle to jump from rest to 9m/s instantly without any time gap, that would require an acceleration close to infinity which is not possible.

3) Area under the curve will give us the total displacement which will be 9×6 = 54m