A car starts from rest and travels for 6.0s with a uniform acceleration of 2.7m/s ^2. The driver then applies the brakes, causing a uniform acceleration of -2.0m/s^2. If the brakes are applied for 2.0s, determine each of the following.

a.) how fast is the car going at the end of the braking period?

b.) how far has the car gone?

I believe 486 m should be 48.6 m

After six seconds, the car attains a velocity of 6*2.7 = 16.2 m/s, and will; have travelled a distance (a/2)*t^2 = 486 m.

Then the deceleration starts.

(a) After two seconds of deceleration,
V = 16.2 - 2.0*2 = 12.2 m/s

(b) Average speed during deceleration for 2 seconds = 14.2 m/s

Total distance travelled = 486 + 2*14.2 = 514.4 m

A car starts from rest and travels for 5.1 s with

a uniform acceleration of 3.6 m/s
2
. The driver
then applies the brakes, causing a uniform
acceleration of −1.3 m/s
2
.
If the brakes are applied for 2.1 s, how fast
is the car going at the end of the braking
period?
Answer in units of m/s.

a.) How fast is the car going at the end of the braking period? Well, I don't want to brake your spirits, but let's find out! We can use the formula v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time.

In this case, the initial velocity is the velocity at the end of the first 6.0 seconds. We can find that using v = u + at. Since we know the uniform acceleration is 2.7 m/s^2 and the time is 6.0 seconds, we can plug in the values and calculate it. But hey, don't worry, I'll do the math for you!
v = 0 + (2.7 m/s^2)(6.0 s) = 16.2 m/s
So, the initial velocity at the end of the first 6.0 seconds is 16.2 m/s. Now we can find the final velocity by using the same formula, but with a negative acceleration this time. The brake causes a uniform acceleration of -2.0 m/s^2 for 2.0 seconds.
v = 16.2 m/s + (-2.0 m/s^2)(2.0 s)
v = 16.2 m/s - 4.0 m/s = 12.2 m/s
So, the car is going at 12.2 m/s at the end of the braking period. Phew, hope it didn't brake your expectations!

b.) How far has the car gone? To find that out, we can use the formula s = ut + (1/2)at^2, where s is the distance traveled, u is the initial velocity, a is the acceleration, and t is the time.
First, let's find the distance traveled during the first 6.0 seconds of uniform acceleration using this formula. Since the car starts from rest, the initial velocity u is 0 m/s, and the time t is 6.0 seconds.
s = (0 m/s)(6.0 s) + (1/2)(2.7 m/s^2)(6.0 s)^2 = 0 m + 0.5(2.7 m/s^2)(36.0 s^2) = 0 + 0.5(2.7)(36) m = 54.0 m
So, the car has traveled 54.0 meters during the first 6.0 seconds. Now let's find out how far it goes during the 2.0 seconds of braking.
To calculate the distance, we use the same formula, but with the final velocity v as 12.2 m/s and the acceleration a as -2.0 m/s^2.
s = (12.2 m/s)(2.0 s) + (1/2)(-2.0 m/s^2)(2.0 s)^2 = 24.4 m + 0.5(-2.0 m/s^2)(4.0 s^2) = 24.4 m + 0.5(-2)(4) m = 24.4 m - 4.0 m = 20.4 m
So, the car goes an additional 20.4 meters during the 2.0 seconds of braking. Adding up the distances, the car has gone a total of 54.0 m + 20.4 m = 74.4 meters. Hooray, it's gone quite a distance, hasn't it?

To solve this problem, we can use the equations of motion that relate acceleration, time, initial velocity, and displacement.

Let's start by find the car's velocity at the end of the acceleration period.

a.) To find the velocity at the end of the acceleration period, we can use the equation:
v = u + at

where:
v = final velocity
u = initial velocity
a = acceleration
t = time

Before the braking, the car was at rest. So, the initial velocity (u) is 0 m/s. The acceleration during this phase is 2.7 m/s^2, and the time (t) is 6.0 s.

Plugging the values into the equation, we have:
v = 0 + (2.7)(6.0)
v = 16.2 m/s

Therefore, at the end of the acceleration period, the car is moving at a velocity of 16.2 m/s.

b.) To find the distance traveled during the acceleration period, we can use the equation:
s = ut + (1/2)at^2

where:
s = distance
u = initial velocity
a = acceleration
t = time

Again, the initial velocity (u) is 0 m/s, the acceleration (a) is 2.7 m/s^2, and the time (t) is 6.0 s.

Plugging the values into the equation, we have:
s = (0)(6.0) + (1/2)(2.7)(6.0)^2
s = 0 + (1/2)(2.7)(36)
s = 0 + (1/2)(97.2)
s = 48.6 m

Therefore, during the acceleration phase, the car travels a distance of 48.6 meters.

Now, let's move on to the braking phase.

During the braking phase, the acceleration is -2.0 m/s^2, and the time is 2.0 s.

a.) To find the final velocity at the end of the braking, we can use the same equation as before:
v = u + at

However, in this case, our initial velocity is the final velocity from the acceleration phase, which is 16.2 m/s. The acceleration (a) is -2.0 m/s^2, and the time (t) is 2.0 s.

Plugging the values into the equation, we have:
v = 16.2 + (-2.0)(2.0)
v = 16.2 - 4
v = 12.2 m/s

Therefore, at the end of the braking period, the car is moving at a velocity of 12.2 m/s.

b.) To find the distance traveled during the braking period, we can use the same equation as before:
s = ut + (1/2)at^2

Here, the initial velocity (u) is the final velocity from the acceleration phase, which is 16.2 m/s. The acceleration (a) is -2.0 m/s^2, and the time (t) is 2.0 s.

Plugging the values into the equation, we have:
s = (16.2)(2.0) + (1/2)(-2.0)(2.0)^2
s = 32.4 - (1/2)(2.0)(4)
s = 32.4 - 4
s = 28.4 m

Therefore, during the braking phase, the car travels a distance of 28.4 meters.

To summarize:
a.) The car is moving at a velocity of 12.2 m/s at the end of the braking period.
b.) The car travels a distance of 28.4 meters during the braking phase.