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 3.5 m/s2 for 3.6 seconds. It then continues at a constant speed for 10.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 167 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.

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 3.5 m/s2 for 3.6 seconds. It then continues at a constant speed for 10.1 seconds, before applying the brakes such that the car’s speed decreases uniformly coming to rest 167 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.

How far does the blue car travel before its brakes are applied to slow down?

Answer 1

Add the distance travelled by the blue car while accelerating for 3.6 s to the distance travelled at constant speed for 10.1 s. You don't need the information about the yellow car or the stopping distance.

Answer 2

okay well I got stuck, my new question is how do I find the distance while the car is accelerating at the 3.5 m/s^2 for 3.6 seconds?

Answer 3

that is easy actually.....

d(distance)= 1/2at^2 +v(initial)t, where a-acceleration, t-time. v-initial here is 0 because the car starts from rest....so d= 1/2at^2 which is .5(3.5)(3.6^2)...which comes out to 22.68m I believe

Answer 4

To find the distance traveled by the blue car before its brakes are applied to slow down, we need to break down the motion of the car into different segments and calculate the distance traveled in each segment.

Segment 1: Acceleration phase
The blue car accelerates uniformly at a rate of 3.5 m/s^2 for 3.6 seconds. To find the distance traveled during this phase, we can use the formula:

distance = initial velocity * time + 0.5 * acceleration * time^2

At the beginning of this phase, the initial velocity is 0 m/s since the car was at rest. Plugging in the values, we get:

distance1 = 0 * 3.6 + 0.5 * 3.5 * (3.6)^2
distance1 = 0 + 0.5 * 3.5 * 12.96
distance1 = 0 + 0.5 * 3.5 * 12.96
distance1 = 22.68 meters

Segment 2: Constant speed phase
After accelerating, the blue car continues at a constant speed for 10.1 seconds. The distance traveled during this phase can be calculated by multiplying the constant speed by the time:

distance2 = constant speed * time
distance2 = constant speed * 10.1

Since the distance traveled in this phase is not given, we cannot calculate it precisely without additional information.

Segment 3: Deceleration phase
The blue car applies the brakes and decelerates uniformly to come to a stop. The total distance traveled during this phase is given as 167 meters.

Segment 1 and 2 distance = distance1 + distance2 = 22.68 + constant speed * 10.1

Total distance = distance1 + distance2 + 167 = 22.68 + constant speed * 10.1 + 167

To find the constant speed during segment 2, we need to know the distance traveled during this phase or additional information about the speed. Without this information, we cannot determine the exact distance traveled by the blue car before its brakes are applied to slow down.