A car traveling initially at +7.7 m/s accelerates uniformly at the rate of +0.76 m/s^2 for a distance of 233m.
- what is its velocity at the end of acceleration (answer m/s)

Question

anonymous · Answer

Known values are v0 = 7.7m/s, a = 0.76 m/s*s, d = 233. 
The final velocity (v) apparently depends on time (t) and constant acceleration (a) as 

v = v0 + a*t;                                   (1)

From the other side, we know that the distance (d) related with constant acceleration, velocity, and time by an equation 

d = v0*t+a*t^2/2.                          (2)

So, we can immediately find the t (quadratic equation):

t = (-v0+_(v0^2+4ad)^0.5)/a.

My calculation gives such answers t1 =  16.62 s, t2 = -36.89 s. The second one seems meaningless, so t = 16,62 s.
Substitution that in equation (1) gives v = 20,33 m/s.

anonymous · Answer

what would it be for 117 m

anonymous · Answer

Oh, seems I miscalculated above. When distance is 233 m t is 26.32 s and v = 7.7 + 0.76*26.32 = 27.70 m/s. When it is 117 m I got t = 16.67 s and v = 7.7 + 0.76*16.67 = 20.37. That's if I didn't miscalculated again.