A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the
particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s?
The accelerationtime graph looks like a straight line and the slope is negative and it makes a triangle with base of 6 units and a height of 6 units.
anonymous
 one year ago
At t = 0, a particle is located at x = 25 m and has a velocity of 15 m/s in the positive x direction. The acceleration of the particle varies with time as shown in the diagram. What is the position of the particle at t = 5.0 s? The accelerationtime graph looks like a straight line and the slope is negative and it makes a triangle with base of 6 units and a height of 6 units.

This Question is Closed

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1dw:1444427089918:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1if that drawing is correct, you have \(a(t) = 6t\) you know that velocity \(v(t) = \int a(t) \; dt \; [ [A]]\) you know that displacement/position is given by \(x(t) = \int \; v(t) \; dt\; [ [B] ]\) you also have boundary conditions: \( x(0) = 25 m, v(0) = +15 m/s\) integrate [A] then [B], and then use the conditions to solve for the constants then solve for x(5).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry for taking so long to replay but the site was down for me for some reason. That worked perfectly i got the equation from the graph and integrated it twice like you said and i got the correct answer, so thank you!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.