A car which is traveling at a velocity of 9.6 m/s undergoes an acceleration of 4.2 m/s^2 over a distance of 450 m. How fast is it going after acceleration?
+33661 You can use \(v^2=u^2+2as\)
v = final velocity
u = initial velocity
a = acceleration
s = distance
\(v=\sqrt{9.6^2+2*4.2*450}=62.2 m/s\)
+130513 I'm not great at Physics, but I think we can use this equation to solve this:
[vf ]2 = [vi]2 + 2ad where vf is the final velocity, vi is the intal velocity, a is the acceleration and d is the displacement of the object.......so we have
[vf]2 = (9.6m's)2 + 2(4.2m/s2)(450m)
[vf]2 = 3872.16m2/s2 taking the square root of both sides we have
vf = about 62.23 m/s
