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.

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

14.01 m/s

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

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

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

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.

\[a = (vi-vf)/(t)\] 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.

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...:)

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

plz correct this...:)

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