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 3.7 m/s2 for 4.4 seconds. it then continues at a constant speed for 8.3 seconds, before applying the brakes such that the carâs speed decreases uniformly coming to rest 216.0 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. 1. how fast is the blue car going 1.8 seconds after it starts? 2. how fast is the blue car going 10.0 seconds after it starts? 3. how far does the blue car travel before its brakes are applied to slow down? 4. what is the acceleration of the blue car once the brakes are applied? 5. what is the total time the blue car is moving? 6. what is the acceleration of the yellow car?

1. How fast is the blue car going 1.8 seconds after it starts?

Recall this kinematic equation:

Vf = Vi + aΔt

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest)

a = 3.7 m/s² (this is the acceleration between t = 0s and t = 4.4s)

Δt = 1.8 s

Substitute the terms in the equation with the given values and solve for Vf:

Vf = 0 + 3.7×1.8

2. How fast is the blue car going 10.0 seconds after it starts?

The car stops accelerating after t = 4.4s and continues at a constant velocity for the next 8.3 seconds. This means the car is traveling at a constant velocity between t = 4.4s and t = 12.7s. At t = 10s the car is still traveling at this constant velocity.

We must use the kinematic equation from the previous question to solve for this velocity. Use the same values except Δt = 4.4s which is the entire time interval during which the car is accelerating:

Vf = 0 + 3.7×4.4

Vf = 16.28 m/s

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

We must break down the car's path into two parts: When it is traveling under constant acceleration and when it is traveling at constant velocity.

Traveling under constant acceleration:

Recall this kinematic equation:

d = ×Δt

d is the distance traveled.

Vi is the initial velocity.

Vf is the final velocity.

Δt is the amount of elapsed time.

Given values:

Vi = 0 m/s (the car starts at rest).

Vf = 16.28 m/s (determined from question 2).

Δt = 4.4 s

Substitute the terms in the equation with the given values and solve for d:

d = ×4.4

d = 35.8 m

Traveling at constant velocity:

Recall the relationship between velocity and distance:

d = vΔt

d is the distance traveled.

v is the velocity.

Δt is the amount of elapsed time.

Given values:

v = 16.28 m/s (the constant velocity from question 2).

Δt = 8.3 s (the time interval during which the car travels at constant velocity)

Substitute the terms in the equation with the given values:

d = 16.28×8.3

d = 135.1 m

Add up the distances traveled.

d = 35.8 + 135.1

4. What is the acceleration of the blue car once the brakes are applied?

Recall this kinematic equation:

Vf²=Vi²+2ad

Vf is the final velocity.

Vi is the initial velocity.

a is the acceleration

d is the distance traveled.

Given values:

Vi = 16.28 m/s

Vf = 0 m/s

d = 216 m - 170.9 m = 45.1 m (subtracting the distance already traveled from the total path length)

Substitute the terms in the equation with the given values and solve for a:

0² = 16.28²+2a×45.1

5. What is the total time the blue car is moving?

We already know the time during which the car is traveling under constant acceleration and traveling at constant velocity. We now need to solve for the amount of time during which the car is decelerating.

Recall again:

d = ×Δt

Given values:

d = 45.1 m

Vi = 16.28 m/s (the velocity the car was traveling at before hitting the brakes).

Vf = 0 m/s (the car slows to a stop).

Substitute the terms in the equation with the given values and solve for Δt:

45.1 = ×Δt

Δt = 5.54s

Add up the times to get the total travel time:

t = 4.4 + 8.3 + 5.54 =

6. What is the acceleration of the yellow car?

Recall this kinematic equation:

d = ViΔt + 0.5aΔt²

d is the distance traveled.

Vi is the initial velocity.

a is the acceleration.

Δt is the amount of elapsed time.

Given values:

d = 216 m (both cars meet at 216m)

Vi = 0 m/s (the car starts at rest)

Δt = 18.24 s (take the same amount of time to reach 216m)

Substitute the terms in the equation with the given values and solve for a:

216 = 0×18.24 + 0.5a×18.24²

this is an example of "committed" time.

committed time, as contracted time, takes need over essential and available time since it is seen as profitable work. it alludes to the time allotted to keep up a home and family. at the point when a man is commuting using committed time this individual may feel that the drive is an obligation to family, for example, strolling youngsters to class or driving a mate to work. contracted and committed time clients may feel that their commute could really compare to the commute of important or leisure time clients on the grounds that their commute is profitable work. in this manner, they might be more disposed to pick a mechanized method of movement.  

365.25 earth days are in the year of a light year