A community for students.
Here's the question you clicked on:
 0 viewing
AJTurnbull
 4 years ago
a car is traveling in a straight line and decelerating at a constant rate. after 3.2 s it traveled 88m and slowed to a speed of 17m/s what was the initial speed of the car?
AJTurnbull
 4 years ago
a car is traveling in a straight line and decelerating at a constant rate. after 3.2 s it traveled 88m and slowed to a speed of 17m/s what was the initial speed of the car?

This Question is Closed

nikvist
 4 years ago
Best ResponseYou've already chosen the best response.0\[t=3.2s\quad,\quad s=88m\quad,\quad v=17m/s\]\[v=v_0at\quad\Rightarrow\quad a=\frac{v_0v}{t}\]\[v^2=v_0^22as=v_0^22\frac{v_0v}{t}s=v_0^2\frac{2s}{t}v_0+\frac{2s}{t}v\]\[v_0^2\frac{2s}{t}v_0+\frac{2s}{t}vv^2=0\]\[v_0=\frac{\frac{2s}{t}\pm\sqrt{\left(\frac{2s}{t}\right)^24\left(\frac{2s}{t}vv^2\right)}}{2}=\]\[=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}\right)^2\frac{2s}{t}v+v^2}=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}v\right)^2}=\]\[=\frac{s}{t}\pm\left(\frac{s}{t}v\right)\quad\mbox{(only sign +)}\]\[v_0=\frac{2s}{t}v=38m/s\]

shubham
 4 years ago
Best ResponseYou've already chosen the best response.2OR Use this equation of motion S = {(u+v)/2}t 88m = {(u+17)/2}3.2 u=5513 = 38m/s

shubham
 4 years ago
Best ResponseYou've already chosen the best response.2For constant accelarations, velocity is linear function of time and hence you can assume mean of initial and final velocities to be the average velocity. So, you don't need a actually.

AJTurnbull
 4 years ago
Best ResponseYou've already chosen the best response.0oh that makes it easy been trying to figure that out all night
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.