A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 one year ago
you drive with a constant speed of 13.5 m/s for 30 s. you then accelerate for 10s to a speed of 22 m/s you then slow to a stop in 10s how far have you traveled?
anonymous
 one year ago
you drive with a constant speed of 13.5 m/s for 30 s. you then accelerate for 10s to a speed of 22 m/s you then slow to a stop in 10s how far have you traveled?

This Question is Open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0case I when costt. v= 13.5 m/s time taken,t = 30 s dist. covered,\[s_{1}\]= v *t = 13.5 * 30 =405 m case II initial vel. , u = 13.5 m/s final vel. , v= 22 m/s time taken ,\[ t _{2}=10 s\] thus acc. ,a = (vu)/t = (22 13.5)/10 = .085 m/s^2 hence, \[v ^{2}u^{2}=2as\] \[22^{2}13.5^{2}= 2(0.85)*s\] 484182.25 = 1.7 *s \[s _{2}=177.5 m\] caseIII v=0 u=22 m/s t=10 s a=(vu)/t = 2.2m/s^2 \[v ^{2}u ^{2}=2as\] 22 *22 /[2 *(2.2)] =s \[s_{3}=110 m\] total dist.= \[s _{1}+s _{2}+s _{3}= \] 405 + 177.5 + 110 = 692.5 m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444222868559:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the area under the graph gives the distance covered . this is 2nd method to calculate the distance traveled. area = area(sq.) + ar(trapezium) + ar(triangle) u get the 3 equations in a single step
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.