Touble with acceleration/velocity problem

[Hwarang]

trouble with acceleration/velocity problem

The land speed record of 13.9 m/s (31 mi/h) for birds is held by the Australian emu. An emu running due south in a straight line at this speed slows down to a speed of 12.0 m/s in 3.3 s.

Assuming that the acceleration remains the same, what is the bird's velocity after an additional 3.7 s has elapsed?


i know that you have to use...


A = ((Vfinal - Vinitial)/(Tfinal - Tinitial))


So we're solving for Vfinal.. For Vinitial, it would be 13.9m/s? And A would be -12m/s (neg. because slowing down). And for time.. would it be 7 sec-0 sec.?

thanks

 

[whozum]

You want to find the acceleration from the information given in thefirst two sentnces. Then from there use that acceleration to find the velocity at 7s. 13.9 would be hte initial velocity.

 

[Hwarang]

thx

thx! that helped a lot.
i got -4.0 m/s as the velocity... but i was wondering, can velocity be negative? is it because the emu was slowing down?

 

[whozum]

Was your average acceleration 0.58m/s^2? Thats what I got, then

\Delta v = a t

\Delta v = 0.58 \times 7 = 4.06 m/s

Recall this is the CHANGE in velocity. You started at 13.9m/s, and accelerating at that rate for 7 seconds you will lose 4.06 m/s. Whatis your final velocity?

Velocity CAN be negative, but in this case it shouldn't be.

 

[Hwarang]

sorry.. I'm a bit confused.

if Vinitial =13.9m/s.. and the change in velocity = 4m/s...
by plugging in everything
-0.5758m/s^2 = ((Vfinal - 13.9)/(7sec))
Vfinal = 9.86 m/s

 

[omairt2001]

yep thts rite


y don't u just use

v = u + at

where u=13.9 t=7 and a=-0.576

 

[Hwarang]

thx everyone!

 

[whozum]

Thats the right answer.

y don't u just use

v = u + at

where u=13.9 t=7 and a=-0.576

He did, the correct equation is \Delta v = a\Delta t[/tex]<br /> <br /> By definitions of the delta operator it comes out to<br /> <br /> V_f - V_i = at where v_0 \ and \ v_i [/tex] are analogous.