A car accelerates from rest at 1.70 m/s2 along a 220-m stretch of road. Find the car’s velocity at the end of the stretch.

Answer 1

d = 0.5at^2,
220 = 0.5 * 1.7 * t^2,
220 = 0.85t^2,
t^2 = 220 / 0.85 = 258.8,
t = sqrt(258.8) = 16.1 m.

V = at,
= 1.7 * 16.1 = 27.3 m/s.

Answer 2

CORRECTION:t = sqrt(258.8) = 16.1 s.