Help with a physics question due Tomorrow!!!!!?

Question

A car starts from rest and travels for 4.9 s with a uniform acceleration of +1.2 m/s2. The driver then applies the brakes, causing a uniform acceleration of -2.2 m/s2. The breaks are applied for 1.70 s.

(b) How far has the car gone from its start?

Answer 1

assumong the starting speed of the car is 0.

first motion:
accaleration: 4,9 * 1,2 = endspeed of 5,88 m/s
distance traveled: 0,5 * 4,9 * 5,88 = 14,406 m

second motion:
desceleration 1,7 * -2,2 = -3,74 m/s
distance traveled: 0,5 * 1,7 * 3,74 m/s + (5,88 - 3,74) * 1,7 = 6,817

total motion: 14,406 +6,817 = 21,223 m

Answer 2

Leg1:
s = (1/2)at^2
s = (1/2)(1.2)(4.9)^2
s = 14.406
v = at
v = (1.2)(4.9)
v = 5.88m/s
Leg2:
s = v(zero)t + (1/2)at^2
s = (5.88)(1.70) + (1/2)(-2.2)(1.70)^2
s = 6.817m
Total:
s(total) = 14.401 + 6,817 = 21.218m