Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.4 m/s^2 for 4.1 seconds.

Question

Two cars start from rest at a red stop light. When the light turns green, both cars accelerate forward. The blue car accelerates uniformly at a rate of 3.4 m/s² for 4.1 seconds. It then continues at a constant speed for 14.8 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 260.61 meters from where it started. The yellow car accelerates uniformly for the entire distance, finally catching the blue car just as the blue car comes to a stop.

What is the acceleration of the yellow car?

Answer 1

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

x = 1/2at² = 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 = 1/2at² or 2x/t² = a

a = (2•260.61)/22.59²

a = 1.0214 m/s²

Rounding to 1 sig fig gives a = 1.0 m/s²