anonymous
  • anonymous
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?
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
nikvist
  • nikvist
\[t=3.2s\quad,\quad s=88m\quad,\quad v=17m/s\]\[v=v_0-at\quad\Rightarrow\quad a=\frac{v_0-v}{t}\]\[v^2=v_0^2-2as=v_0^2-2\frac{v_0-v}{t}s=v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v\]\[v_0^2-\frac{2s}{t}v_0+\frac{2s}{t}v-v^2=0\]\[v_0=\frac{\frac{2s}{t}\pm\sqrt{\left(\frac{2s}{t}\right)^2-4\left(\frac{2s}{t}v-v^2\right)}}{2}=\]\[=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}\right)^2-\frac{2s}{t}v+v^2}=\frac{s}{t}\pm\sqrt{\left(\frac{s}{t}-v\right)^2}=\]\[=\frac{s}{t}\pm\left(\frac{s}{t}-v\right)\quad\mbox{(only sign +)}\]\[v_0=\frac{2s}{t}-v=38m/s\]
anonymous
  • anonymous
OR Use this equation of motion S = {(u+v)/2}t 88m = {(u+17)/2}3.2 u=55-13 = 38m/s
anonymous
  • anonymous
how did you get a?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
For 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.
anonymous
  • anonymous
oh that makes it easy been trying to figure that out all night

Looking for something else?

Not the answer you are looking for? Search for more explanations.