During the acceleration of the blue car, it travels a distance based on the kinematic equation

x = 1/2at^{2} = 1/2(3.4)(4.1)^{2} = 28.577 m

The velocity attained by the blue car can be calculated from the kinematic equation

v = at

v = (3.4)(4.1) = 13.94 m/s

During that time of constant velocity it travels a distance based on the kinematic equation

x = vt = (13.94)(14.8) = 206.312 m. therefore the total distance traveled to that point by the blue car is

28.577 + 206.312 = 234.889 m.

Since the total distance traveled at the finish is 260.61, then the distance decelerating is

260.61 - 234.889 = 25.721 m.

The time that the blue car spent in that last deceleration can be calculated from the kinematic equation

x = 1/2(v_{f} + v_{i})t

25.721 = 1/2(0 + 13.94)t

25.721 = 6.97t

t = 3.69 s

The total time the blue car was in motion was then 4.1 + 14.8 + 3.69 = 22.59 s

So the yellow car was also in motion for 22.59 s

The acceleration of the yellow car can be calculated from the kinematic equation

x = 1/2at^{2} or 2x/t^{2} = a

a = (2•260.61)/22.59^{2}

a = 1.0214 m/s^{2}

Rounding to 1 sig fig gives a = 1.0 m/s^{2}

Davinder S.

it's actually going to be 1.02. thanks so much for the explanation sir29d