A truck covers 40.0 m in 7.10 s while smoothly slowing down to a final velocity of 2.75 m/s.
(a) Find the truck's original speed.

Question

anonymous · Answer

40=.5*((ia-2.75)/7.1)*7.1^2 solve for ia :)

anonymous · Answer

14.01  m/s

anonymous · Answer

use this equation 

d=.5*a*t^2

with a =  (vi-vf)/(t)

stormfire1 · Answer

That's not correct...I think you got your Vf and Vi mixed up there :)

$$d=\frac{V_i+V_f}{2}t$$$$40m=\frac{V_i+2.75 m/s}{2}7.1s$$$$V_i=8.52m/s$$

anonymous · Answer

but if it said steady  slow down you must use the accelaration one idk tho lol.

stormfire1 · Answer

It's constant acceleration...so you can just use the average velocity/time=distance.  

If you look at your equations you mixed up Vf and Vi if I'm not mistaken.

stormfire1 · Answer

$$a = (vi-vf)/(t)$$ should be $$a = (vf-vi)/(t)=-0.813m/s^2$$Plug that and 8.52 m/s into the equation you used $$d=V_it+\frac{1}{2}at^2$$ you end up with 40m...which confirms the answer.

anonymous · Answer

V = Vo + at

2.75= Vo + at

at = 2.75 - Vo

S = Vo t + ½ a t² = Vo t + ½ (a t) t = 7.10Vo + ½ (2.75 - Vo)7.10 = 40m
7.10Vo-3.55Vo=30.23

3.55Vo=30.23
Vo=30.23/3.55
 Vo= 8.51

now a= vf-vo/t =2.75-8.51/7.10
a=-0.81
 now for  original speed formula is v=vo+at
where u have calculated vo=8.51
                                    a=-o.81
                                    t=7.10...:)

anonymous · Answer

sorry i have done i thing wrong,,,,its 7.10 squared  where i use formula s=vot+atsquare

anonymous · Answer

plz correct this...:)