help in physics please?

Question

An automobile traveling along a straight road
increases its speed from 45 ft/s to 59 ft/s in
180 ft.
If the acceleration is constant, how much
time elapses while the auto moves the 180 ft?
Answer in units of s.

Answer 1

Initial speed 45 ft/s.
Final speed 59 ft/s.
Distance covered 180 ft.

Average speed is (45 + 59) /2 = 52ft/s.
Average speed is also = distance / time. = 180 /t

Or Time = distance / average speed = 180 / 52 = 3.462s.

Acceleration is (change in speed / time) = (59 -45) / 3.462 = 4 ft/ s^2.

Answer 2

Hmmmm

45 to 59 > 52 ft/s avg over 180 ft.
180/52 = 3.46 seconds

Answer 3

I think it is the following:

X = Xinit +1/2(Vinit + Vx)t

180 = 0 + .5*(45+59)t

t=180/(0.5*(45+59))

t = 3.46 seconds

Answer 4

Well first we state d = vt

which can be rewritten as t = d / v

our change in distance is 180 ft and our change in velocity is 14 ft/s

so t = 180 / 14

t = 12.86 seconds

EDIT:
Ok my bad, I'm an idiot. Listen to the guy above me, he's got it right. There are a few things i forgot to take into account.

Oh ok I see what it is. You are supposed to take the average velocity, not the change in velocity.

Answer 5

BKC's solution looks good/OK. If you're asked what the acceleration is during the distance/time, it's 4.044 ft/sec squared

Answer 6

Man I should know this, but I don't.