course Phy 241
72) A police car is traveling in a straight line with constant speed vp. A truck traveling in the same direction with speed 3/2vp passes the police car. The truck driver realizes that she is speeding, and immediately slows down at a constant rate until she comes to a stop. This is her lucky day, however, and the police car still moving with the same constant speed passes the truck driver without giving her a ticket.
A) show that the truck? speed at the instant that the police car passes the truck does not depend on the magnitude of the truck? acceleration as it slows down, and find the value of the speed.
B) Sketch the x-t graph for the two vehicles.
The point at which the truck passes the police car is only a function of the initial velocity, initial position, and acceleration before the car slows down. At the point when the acceleration changes, the cars have already passed. This can further be demonstrated with a graphical representation of and x-t plot. We can graph one line that begins at a lower x intercept and a low slope, this will represent the police car. We can graph the truck as having a greater x intercept and catching up with the car in a certain amount of time due to the greater slope of this line. The point at which they pass will be represented by the two lines crossing in the x vs t graph. The line that represents the truck will cross the line with the lower slope and then change slope to denote the slowing down.
Graphically, by direct reasoning, or algebraically we can conclude that the amount by which the truck's speed initially exceeds that of the cop is the same as the amount by which the cop's speed will exceed that of the truck at the moment it is passed by the cop.
Graphically, the two v vs. t graphs will form straight lines that intersect at a certain point, corresponding to the instant and velocity at which the two vehicles have the same velocity.
The area between these lines, from the initial point at which the truck passes the cop, forms a triangle representing the extra distance traveled by the truck between passing and coming back down to the speed of the cop.
The lines then diverge, and the subsequent area between the lines represents the distance being made up by the cop. When this area is equal to the area of the former triangle, the cop will pass the truck.
At this instant, the two triangles will be similar. None of this depends on the slope of the truck's v vs. t graph. The similarity of the two triangles shows that the truck's speed when the cop passes it will be as far below the cop's speed as the truck's initial speed was above that of the cop.
The graph requested was of x vs. t. For the cop the graph will be a straight line. For the truck the graph will be part of a parabola, concave downward, intersecting the cop's straight-line graph at a greater slope, then increasing at a decreasing rate until it again intersects the straight-line graph, this time with a lesser slope.
77) A physics student with too much free time drops a water-melon from the roof of a building. He hears the sound of a watermelon going ?plat? 2.50s later. How high is the building? The speed of sound is 340 m/s. You may ignore air resistance.
If we hear the splat 2.50 seconds later we need to think in terms of two v vs t graphs. One represents the fall of the melon and the other represents the speed of the sound. For the first graph we know the acceleration to be 9.8 m/s^2 and we know that `ds can be found by multiplying .5a`dt^2.
Good, but are you thinking here in terms of a v vs. t graph or a position vs. t graph?
Now we know the other equation that is also equal to `ds deals with the average velocity =`ds/`dt, therefore 340 `dt=`ds. Both `dt will not be equal, rather `dt1 + `dt2 = 2.5. Now we can say that 2.5 ?dt1=`dt2 to eliminate a variable. Setting the two equations equal we obtain (.5)(9.8m/s^2)(`dt1^2)=340(2.5-`dt1). We can solve this with the quadratic equation and choose `dt1 as 2.42 seconds. If `dt1 is 2.42 seconds, `dt2 is .08 seconds. Now to find `ds we can simply plug in `dt2 into the equation `ds = .08*340 to yield a `ds of about 28.6 m.
Very good.
82) A model rocket has a constant upward acceleration of 40.0 m/s^2 while its engine is running. The rocket is fired vertically, and the engine runs for 2.5 seconds before it uses up the fuel. After the engine stops, the rocket is in free fall. The motion of the rocket is purely up and down. a) sketch the a vs t, v vs t, y vs t graphs for the rocket. B) What is the maximum height that the rocket reaches? C) What will be the speed of the rocket just before it hits the ground d) is the total time of the fight equal to twice the time the rocket takes to reach its highest point? Why or y not?
The a vs t graph has two segments of horizontal lines because in both situations acceleration is constant, one at 40 m/s^2, and one at -9.8 m/s^2. On the velocity v vs t graph the first segment of the line begins at the origin and has a slope of 40. The first segment stops at 100m/s. The second line has a slope of -9.8. The y vs t is increasing at an increasing rate for the first part and then decreases at an increasing rate.
y vs. t will continue to increase until the acceleration of gravity has slowed it to 0 velocity, which will take over 10 seconds. The graph will not suddenly start decreasing.
Starting at t = 2.5 seconds the graph will form a concave-downward parabola, which at first increases at a decreasing rate until reaching a peak (the vertex of the parabola) then begins decreasing at an increasing rate.
The maximum height the rocket reaches occurs when the velocity is at 100 m/s.
The rocket continues rising for another 10.2 seconds, as its upward velocity decreases to 0
We can find this by showing on the v graph that at v=0, t=0 at v=40, t=0 and so on until t=2.5 v=100. At a velocity of 100 m/s and an average velocity of 50 m/s, the `ds will be 50*2.5 or 125 m. To find the speed right before the rocket hits the ground we can first find how long it will take to hit the ground. Since at the maximum height v=0, we know v0 is 0 so we can use `ds=.5a`dt^2. If a is 9.8 and `ds is 125 we can solve for `dt to be 5.05 seconds. Now we can use a =`dv/`dt and solve for `dv of 49.49 m/s. If it starts at 0 vf is 49.5 m/s. d) It almost is but not quite because total forces acting on the object are different than when it comes down.
Your work on these last few questions will change a bit because of the continuing upward motion.
87) Spiderman steps from the top of a tall building. He freely falls from rest to the ground a distance of h. he falls a distance h/4 in the last 1.0s of his fall. What is the h of the building? H is represented by the equation h=(.5)(9.8)`dt^2. h/4 can be set equal to h-(h/4)+v0(1)+.5(9.8)(1^2). We can set these equal to one another to solve for `dt if we find v0. How can we find v0 from this information, or is there another way to set up this problem?
His position at clock time t is h - .5 g t^2. If `dt is the total time of fall then we know two things:
h - .5 g `dt^2 = 0 and
h - .5 g (`dt-1s)^2 = h/4.
This gives you two simultaneous equations in the variables h and `dt.
90) A student steps of a skyscraper that is 180 m high. Five seconds later superman arrives and begins free falls off the building starting at an initial velocity v0 generated from pushing himself from downward from the edge of the roof with his legs of steel. He then falls with the same acceleration as any freely falling body. What must the value of v0 be so that Superman catches the student just before they reach the ground? B) on the same graph sketch the position of the student and superman as functions of time. Take superman? initial speed to have the value calculated in part a. c) If the height of the skyscraper is less than some minimum value, even Superman cant reach the student before he hits the ground. What is this minimum height?To find the value of v0 we must first develop an equation for the student? free fall. Since he starts out with v0=0 and x=0, then `ds=.5a`dt^2. We can solve for `dt which is about 6.06. If superman leaves 5 seconds later, he only has 1.06 seconds before the student hits the ground. So we set up superman? equation as 180=v0(1.06)+.5(9.8)(1.06^2) Solving for v0, this equals 164 m/s. If the equation is set up in this manner it seems that the minimum height will be 180 m.
Good. This problem is also addressed in the Query.
See my notes and let me know if you have questions.