A car starts from rest and travels for 7.0 s with a uniform acceleration of +2.3 m/s2. The driver then applies the brakes, causing a uniform negative acceleration of -1.0 m/s2. If the brakes are applied for 2.0 s, how fast is the car going at the end of the braking period, and how far has it gone?

break up the question into segments

a. travels for 7.0 s with a uniform acceleration of +2.3 m/s2
b. then applies the brakes, causing a uniform negative acceleration of -1.0 m/s2. If the brakes are applied for 2.0 s,

figure the end velocicty of a, which will be the starting velocicty of b, then find the end velocity.
distance can be found in each, and added.

Oh, braking! That's when cars turn into real drama queens on the road. Alright, let's crunch some numbers!

For the first part, we'll look at the car's acceleration. So, it starts from rest and accelerates with an impressive +2.3 m/s² for 7.0 seconds. That's like a rocket ship on a budget! Anyway, we can find the final velocity of the car during this time by using the formula:

v = u + at

Where:
v = final velocity
u = initial velocity (which is zero in this case)
a = acceleration (which is +2.3 m/s²)
t = time (which is 7.0 seconds)

Using that formula, we find that the car's speed at the end of the acceleration period is 2.3 m/s² × 7.0 s = 16.1 m/s.

Now, let's move on to the exciting part - braking! The car experiences a uniform negative acceleration of -1.0 m/s² for 2.0 seconds. To figure out the final velocity during this period, we can use the same velocity formula:

v = u + at

Where:
v = final velocity
u = initial velocity (which is 16.1 m/s from the acceleration period)
a = acceleration (which is -1.0 m/s²)
t = time (which is 2.0 seconds)

Using that formula, we find that the car's speed at the end of the braking period is (16.1 m/s) + (-1.0 m/s² × 2.0 s) = 14.1 m/s.

Now, let's calculate the distance the car has traveled during the whole ordeal. To do that, we can use the formula:

s = ut + (1/2)at²

Where:
s = distance
u = initial velocity (which is zero because the car starts from rest)
t = time (which is 7.0 seconds for the acceleration period and 2.0 seconds for the braking period)
a = acceleration (which is +2.3 m/s² for the acceleration period and -1.0 m/s² for the braking period)

Let's compute it piece by piece.

For the acceleration period:
s1 = (0 m/s × 7.0 s) + (1/2)(2.3 m/s²)(7.0 s)² = 72.87 m

For the braking period:
s2 = (16.1 m/s × 2.0 s) + (1/2)(-1.0 m/s²)(2.0 s)² = 31.6 m

So, the total distance covered by the car is s1 + s2 = 72.87 m + 31.6 m = 104.47 m.

So, to recap: the car is going at a speed of 14.1 m/s at the end of the braking period, and it has traveled a total distance of 104.47 m. I hope the car enjoyed its little adventure, although I'm sure the driver had a few heart-stopping moments along the way!

To solve the problem, we need to use the equations of motion for both the acceleration period and the braking period.

1. Acceleration period:
Using the equation of motion: v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.
Initial velocity u = 0 m/s (car starts from rest)
Acceleration a = +2.3 m/s^2
Time t = 7.0 s

v = u + at
v = 0 + (2.3)(7.0)
v = 16.1 m/s

Therefore, the speed of the car at the end of the acceleration period is 16.1 m/s.

2. Braking period:
Using the same equation of motion: v = u + at, this time with negative acceleration.
Initial velocity u = 16.1 m/s (speed at the end of acceleration period)
Acceleration a = -1.0 m/s^2 (negative because the car is decelerating)
Time t = 2.0 s

v = u + at
v = 16.1 + (-1.0)(2.0)
v = 16.1 - 2.0
v = 14.1 m/s

Therefore, the final speed of the car at the end of the braking period is 14.1 m/s.

To find the distance traveled during the braking period, we can use the equation: s = ut + (1/2)at^2.
Using the same values as before, except we need to change the initial velocity to 16.1 m/s because that is the speed at the end of the acceleration period.

s = ut + (1/2)at^2
s = 16.1 * 2.0 + (1/2)(-1.0)(2.0)^2
s = 32.2 - 2.0
s = 30.2 m

Therefore, the car has traveled a distance of 30.2 meters during the braking period.

To find the final velocity and the distance traveled by the car, we can break the problem into two parts: the first part where the car is accelerating, and the second part where the car is decelerating.

1. Acceleration phase:
Here, the car starts from rest (initial velocity, u = 0) and travels for 7.0 seconds with a uniform acceleration of +2.3 m/s^2. We need to find the final velocity (v) and the distance traveled (s) during this phase.

To find the final velocity, we can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values:

v = 0 + (2.3 m/s^2)(7.0 s)
v = 16.1 m/s

So the car's velocity at the end of the acceleration phase is 16.1 m/s.

To find the distance traveled, we can use the equation:

s = ut + (1/2)at^2

where s is the distance, u is the initial velocity, a is the acceleration, and t is the time.

Plugging in the values:

s = (0)(7.0 s) + (1/2)(2.3 m/s^2)(7.0 s)^2
s = 115.15 m

So the car travels 115.15 meters during the acceleration phase.

2. Deceleration phase:
Now, the car applies the brakes for 2.0 seconds with a uniform negative acceleration of -1.0 m/s^2. We need to find the final velocity and the distance traveled during this phase.

Using the same equations as before:

v = u + at

Plugging in the values:

v = 16.1 m/s + (-1.0 m/s^2)(2.0 s)
v = 14.1 m/s

So the car's velocity at the end of the deceleration phase is 14.1 m/s.

To find the distance traveled, we again use the equation:

s = ut + (1/2)at^2

Plugging in the values:

s = (16.1 m/s)(2.0 s) + (1/2)(-1.0 m/s^2)(2.0 s)^2
s = 32.2 m - 2.0 m
s = 30.2 m

So the car travels 30.2 meters during the deceleration phase.

To find the car's final velocity and distance traveled overall, we need to add the values from both phases:

Final velocity = 14.1 m/s
Distance traveled = 115.15 m + 30.2 m = 145.35 m

Therefore, at the end of the braking period, the car is going at a speed of 14.1 m/s and has traveled a distance of 145.35 meters.