Calculate (a) the average velocity of the particle

Click For Summary
The discussion focuses on two physics problems involving motion and velocity calculations. For the first problem, the average velocity of a particle can be calculated using the change in position over time, utilizing the given position function x(t) = 50t + 10t^2. The second problem involves determining the necessary acceleration for the Enterprise to avoid colliding with a Klingon ship by setting up equations of motion for both ships. Participants emphasize the importance of understanding average and instantaneous velocity definitions and how to apply them to solve these problems. Overall, the thread provides guidance on approaching motion-related physics problems effectively.
juicev85
Messages
13
Reaction score
0
my professor gave us these two problems to try. I just started this physics class and am a little lost on these problems. any help/ hints would be greatly appreciated.

1) A particle moves along the x-axis according to the equation,
x(t)=50t + 10t^2(squared)
where t is measured in seconds and x is in meters. Calculate (a) the average velocity of the particle during the first 3.0 seconds of its motion, (b) the instantaneous velocity of the particle at t = 3 s, (c) the instantaneous acceleration at t = 3 s. (d) Graph the equation and indicate how the answer to (a) and (b) can be obtained from the plot. (e) While you are at it, graph velocity vs. time as well.

2) The starship enterprise returns from warp drive to ordinary space with a forward speed of 50 km/s. To the crew’s shocking surprise, a Klingon ship is 100 km directly ahead, traveling in the same direction at a measly 20 km/s. Without evasive action, the Enterprise will overtake and collide with the Klingons in just slightly over 3.0 s. The Enterprise’s computers react instantly to brake the ship. What acceleration does the Enterprise need to just barely avoid a collision with the Klingon ship? Assume the acceleration is constant.

Thanks.
 
Physics news on Phys.org
Hints:

v(t) = \frac{d x(t)}{dt}

a(t) = \frac{d v(t)}{dt}
 
You are given a function of time which describes position of your particle. To find its location at any time t, you just evaluate your function at t. So at t=0 you have x(0)= 50(0)+10(0)2 = 0. You can find its position at any time in the same manner, you should be able to find its position at t=3.

What defining have you been given for average velocity? You know the position at t=0, you know the position at t=3 so you have a change in position and a change in time. You should be able to compute the average velocity.

I am assuming that this is a calculus based course, what is the definition of instantaneous velocity? Apply that to your position function, evaluate at t=3.

What is difficult about plotting functions?
 
thanks a lot guys I solved the first question, oh and I didn't mean to include the last part of the first question (I can plot the function just fine). What about the second question, the star trek question. any ideas about that one?
 
Set up 2 separate equations of motion x_1(t) and x_2(t) (i.e. one for each ship), but use the same axis origin. What do you know about x_1 and x_2 at collision?
 
Beams of electrons and protons move parallel to each other in the same direction. They ______. a. attract each other. b. repel each other. c. neither attract nor repel. d. the force of attraction or repulsion depends upon the speed of the beams. This is a previous-year-question of CBSE Board 2023. The answer key marks (b) as the right option. I want to know why we are ignoring Coulomb's force?
Thread 'Struggling to make relation between elastic force and height'
Hello guys this is what I tried so far. I used the UTS to calculate the force it needs when the rope tears. My idea was to make a relationship/ function that would give me the force depending on height. Yeah i couldnt find a way to solve it. I also thought about how I could use hooks law (how it was given to me in my script) with the thought of instead of having two part of a rope id have one singular rope from the middle to the top where I could find the difference in height. But the...
I treat this question as two cases of Doppler effect. (1) When the sound wave travels from bat to moth Speed of sound = 222 x 1.5 = 333 m/s Frequency received by moth: $$f_1=\frac{333+v}{333}\times 222$$ (2) When the sound wave is reflected from moth back to bat Frequency received by bat (moth as source and bat as observer): $$f_2=\frac{333}{333-v}\times f_1$$ $$230.3=\frac{333}{333-v}\times \frac{333+v}{333}\times 222$$ Solving this equation, I get ##v=6.1## m/s but the answer key is...