Difficult physics problems.

In summary: This problem involves vector addition and using trigonometric functions. Your approach of breaking the distances into horizontal and vertical components is correct. To find the magnitude of the displacement, you will need to use the Pythagorean theorem (a^2 + b^2 = c^2) with the horizontal and vertical components as the legs of a right triangle. To find the angle, you can use the inverse tangent function (tan^-1) to find the angle between the displacement and the horizontal direction.
  • #1
Blink691
17
0
**I've done various steps to theses problems and can't seem to figure them out. If someone can either give me steps how to do them or just work them out for me I'd greatly appreciate it. Thanks!

1) An electron with initial velocity v0 = 2.6x10^5m/s enters a region 1.0 cm long where it is electrically accelerated (Figure 2-26). It emerges with velocity v = 4.70x10^6 m/s. What was its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.)

I ended up getting 1.10 E 15 m/s^2 but I'm not sure if that's correct. I used the Vf^2-Vi^2=2ad equation.



2) You are arguing over a cell phone while trailing an unmarked police car by 36 m. Both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.5 s, the police officer begins emergency braking at 5 m/s2.

a) What is the separation between the two cars when your attention finally returns?

b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

I found a similar problem online to this and just redid the work and changed my numbers but I'm stuck on part b. I'm not even sure if what I have for a is correct. My work is shown below:
distance between car and police= 36m speed of cars v=110km/hr =110(5/18) = 30.55m/s
after police car breaks: initial v=20.55m/s, final v= 0m/s, a=-5m/s^2 from V^2-V^2-2as
0^2-30.55^2=2(-5)s s=933.64/10= 93.364 m

velocity of police car after 2.5 secs of breaking v=v+at 30.55+(-5)2.5=18.05 m/s
distance traveled in this time by police is S: (18.05)^2-(30.55)^2-2(-5)s =60.78m
distance traveled by car in 2.5s=velocity x time: 30.55 x 2.5= 76.375m
separation between 2 cars: (36+60.78)-76.375= 20.405m



3) A car is driven east for a distance of 45 km, then north for 30 km, and then in a direction 35° east of north for 22 km. Sketch the vector diagram and determine the (a) magnitude and (b) angle of the car's total displacement from its starting point.

Here I tried to break the 3 distances into horizontal and vertical, which I thought was working but I ended up forgetting about the angle they give and now I'm totally confused. I'm not exactly sure how to find the magnitude nor the total displacement. I know you usually add up the distances to get the displacement but I have a feeling that's not what you do for this problem. Mind showing me the steps to these problems? Thanks
 
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  • #2
Blink691 said:
**I've done various steps to theses problems and can't seem to figure them out. If someone can either give me steps how to do them or just work them out for me I'd greatly appreciate it. Thanks!

1) An electron with initial velocity v0 = 2.6x10^5m/s enters a region 1.0 cm long where it is electrically accelerated (Figure 2-26). It emerges with velocity v = 4.70x10^6 m/s. What was its acceleration, assumed constant? (Such a process occurs in the electron gun in a cathode-ray tube, used in television receivers and oscilloscopes.)

I ended up getting 1.10 E 15 m/s^2 but I'm not sure if that's correct. I used the Vf^2-Vi^2=2ad equation.

That number looks fine. Believe it or not, electrons really can pick up accelerations that huge from fairly modest electric fields. That's because their masses are so tiny. The force involved is about 1.0x10^-15 N, so the electric field only needs to be about 6300 N/C (or V/m), which is pretty mild.


2) You are arguing over a cell phone while trailing an unmarked police car by 36 m. Both your car and the police car are traveling at 110 km/h. Your argument diverts your attention from the police car for 2.5 s (long enough for you to look at the phone and yell, "I won't do that!"). At the beginning of that 2.5 s, the police officer begins emergency braking at 5 m/s2.

a) What is the separation between the two cars when your attention finally returns?

b) Suppose that you take another 0.4 s to realize your danger and begin braking. If you too brake at 5 m/s2, what is your speed when you hit the police car?

I found a similar problem online to this and just redid the work and changed my numbers but I'm stuck on part b. I'm not even sure if what I have for a is correct. My work is shown below:
distance between car and police= 36m speed of cars v=110km/hr =110(5/18) = 30.55m/s
after police car breaks: initial v=20.55m/s, final v= 0m/s, a=-5m/s^2 from V^2-V^2-2as
0^2-30.55^2=2(-5)s s=933.64/10= 93.364 m

velocity of police car after 2.5 secs of breaking v=v+at 30.55+(-5)2.5=18.05 m/s
distance traveled in this time by police is S: (18.05)^2-(30.55)^2-2(-5)s =60.78m
distance traveled by car in 2.5s=velocity x time: 30.55 x 2.5= 76.375m
separation between 2 cars: (36+60.78)-76.375= 20.405m

That's certainly one way to do it (there are a few similar approaches). I got essentially the same thing (allowing for round-off).


3) A car is driven east for a distance of 45 km, then north for 30 km, and then in a direction 35° east of north for 22 km. Sketch the vector diagram and determine the (a) magnitude and (b) angle of the car's total displacement from its starting point.

Here I tried to break the 3 distances into horizontal and vertical, which I thought was working but I ended up forgetting about the angle they give and now I'm totally confused. I'm not exactly sure how to find the magnitude nor the total displacement. I know you usually add up the distances to get the displacement but I have a feeling that's not what you do for this problem. Mind showing me the steps to these problems? Thanks

Draw a sketch of what the car's path looks like. You'll want to deal with the components of displacement first. How far north of the starting point in total has the car traveled? How far east in total? The relatively hard part is getting the components of that 35º east of north leg (for which you'll need a little trig).
 
  • #3



For the first problem, your answer of 1.10 x 10^15 m/s^2 is correct. You used the correct equation, which is the kinematic equation for constant acceleration. Your approach is correct and it's always a good idea to double check your answer by plugging it back into the equation and making sure it works out. In this case, your answer does work out, so you can be confident that it is correct.

For the second problem, your work is correct up until finding the separation between the two cars. However, for part b, you need to use the equation v^2 = v0^2 + 2aΔx, where v0 is your initial velocity and Δx is the distance traveled during the time you took to realize your danger and begin braking (0.4 seconds). Your initial velocity will be the final velocity of the police car before it began braking (18.05 m/s). Plugging in these values, you should get a final velocity of 2.85 m/s for your car when it hits the police car.

For the third problem, you are on the right track by breaking down the distances into horizontal and vertical components. To find the magnitude of the total displacement, you will use the Pythagorean theorem (a^2 + b^2 = c^2), where a and b are the horizontal and vertical components, respectively, and c is the magnitude of the total displacement. To find the angle, you will use the inverse tangent function (tan^-1) to find the angle between the total displacement and the horizontal direction. Remember to use the correct units for the angle (degrees or radians).

Overall, it's important to double check your work and make sure you are using the correct equations and units. It's also helpful to draw diagrams to visualize the problems and make sure you understand what is being asked. Keep practicing and don't be afraid to ask for help if you are stuck. Good luck!
 

1. What makes a physics problem difficult?

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Improving your problem-solving skills and understanding of fundamental physics concepts can help you tackle difficult problems more effectively. Additionally, practicing with a variety of problems and seeking help from teachers, tutors, or peers can also improve your problem-solving abilities.

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