A car starts from rest and travels for 4.1 s with a uniform acceleration of +1.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.8 m/s2. The breaks are applied for 1.80 s.
I guess you want the total distance. Since v = at and s = vt + 1/2 at^2,
s = (1/2)(1.4)(4.1)^2 + (4.1)(1.4)(1.8) + (1/2)(-1.8)(1.8)^2 = 19.18 m
final velocity is
v = 1.4*4.1 + (-1.8)*1.8 = 2.5 m/s
average speed is 19.18m/(4.1+1.8)s = 3.25 m/s
Well, this car seems to be going through quite an emotional roller coaster! Let's see if we can make sense of it all.
First, the car starts from rest and travels with a positive acceleration of +1.4 m/s2 for 4.1 seconds. This means it's picking up speed, maybe it's in a hurry to get to a clown convention or something.
But wait, the driver then suddenly decides to apply the brakes. Talk about mood swings! The car decelerates with a uniform acceleration of -1.8 m/s2 for 1.80 seconds.
Now, to figure out what happened during these events, we first need to determine the distances traveled during each phase. Are you ready for some serious math clowning?
During the first phase of acceleration, we can use the equation:
distance = initial velocity * time + (1/2) * acceleration * time^2
Since the car starts from rest, the initial velocity is 0. So, the distance traveled during the acceleration phase is:
distance = 0 * 4.1 + (1/2) * 1.4 * (4.1)^2
distance = 0 + 2.8 * 16.81
distance = 46.808 meters
Okay, now for the deceleration phase, we use the same equation, but with the negative acceleration:
distance = 0 * 1.80 + (1/2) * (-1.8) * (1.80)^2
distance = 0 - 0.9 * 3.24
distance = -2.916 meters
Uh-oh, it looks like the car traveled backwards during the braking phase! Maybe it saw a clown chasing after it and got a little scared.
So, to find the total distance traveled, we simply add the distances from each phase:
total distance = 46.808 + (-2.916)
total distance = 43.892 meters
And there you have it, the car traveled a total distance of 43.892 meters during this wild ride. Just another day in the life of a clown car, I suppose!
To solve this problem, we can break it down into two parts: the period of acceleration and the period of braking.
1. Period of acceleration:
We are given the initial velocity, which is 0 m/s, and the acceleration, which is +1.4 m/s^2. We need to find the distance traveled during this period.
We can use the equation:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
Plugging in the values, we get:
distance = (0 * 4.1) + (0.5 * 1.4 * (4.1)^2)
distance = 0 + (0.5 * 1.4 * 16.81)
distance = 0 + 11.7656
distance = 11.77 meters
So, the distance traveled during the period of acceleration is 11.77 meters.
2. Period of braking:
Now, the car is moving with a velocity obtained from the acceleration period. We need to find the distance traveled during this period using the same equation, but with a negative acceleration of -1.8 m/s^2 and a time of 1.80 seconds.
distance = (initial velocity * time) + (0.5 * acceleration * time^2)
distance = (1.4 * 1.80) + (0.5 * -1.8 * (1.80)^2)
distance = 2.52 - 0.5 * 1.8 * 3.24
distance = 2.52 - 0.5 * 5.832
distance = 2.52 - 2.916
distance = -0.396 meters
The negative sign indicates that the car is moving in the opposite direction during this period. Therefore, the car traveled a distance of 0.396 meters in the opposite direction during the period of braking.
Overall, the total distance traveled by the car is the sum of the distances traveled during the acceleration and braking periods:
total distance = distance during acceleration + distance during braking
total distance = 11.77 + (-0.396)
total distance = 11.374 meters
Therefore, the car traveled a total distance of 11.374 meters.
To solve this problem, we need to first find the distance traveled by the car during each phase: the acceleration phase and the braking phase.
During the acceleration phase, the initial velocity is 0 m/s (since the car starts from rest) and the acceleration is +1.4 m/s^2. We can use the kinematic equation:
s = ut + 0.5at^2
where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
Substituting the given values:
s1 = 0 * 4.1 + 0.5 * 1.4 * (4.1^2)
s1 = 0 + 0.5 * 1.4 * 16.81
s1 = 0 + 0.5 * 23.53
s1 = 0 + 11.765
s1 = 11.765 m
So, during the acceleration phase, the car travels 11.765 m.
During the braking phase, the initial velocity is the final velocity from the acceleration phase (which we can find using the equation:
v = u + at
where v is the final velocity, a is the acceleration, t is the time, and u is the initial velocity).
During the acceleration phase, the final velocity is given by:
v1 = u + a1t1
v1 = 0 + 1.4 * 4.1
v1 = 0 + 5.74
v1 = 5.74 m/s
The initial velocity for the braking phase is the final velocity from the acceleration phase, which is 5.74 m/s.
Using the same kinematic equation as before, but with the acceleration during the braking phase (-1.8 m/s^2), we can find the distance traveled during the braking phase:
s2 = u2t2 + 0.5a2t2^2
s2 = 5.74 * 1.8 + 0.5 * (-1.8) * (1.8^2)
s2 = 10.332 - 0.5 * 1.8 * 3.24
s2 = 10.332 - 2.916
s2 = 7.416 m
So, during the braking phase, the car travels a distance of 7.416 m.
The total distance traveled by the car is the sum of the distance traveled during the acceleration phase and the distance traveled during the braking phase:
Total distance = s1 + s2
Total distance = 11.765 + 7.416
Total distance = 19.181 m
Therefore, the car travels a total distance of 19.181 m during the given time intervals.