Pr 2. A car starts from rest and travels for 10.0 s with a uniform acceleration of +1.6 m/s2. The driver then applies the brakes, causing a uniform negative acceleration of -3.0 m/s2. If the brakes are applied for 1.0 s, how fast is the car going at the end of the braking period, and how far has it gone?

To solve this problem, we'll need to break it down into two parts: the initial acceleration phase and the deceleration phase.

First, let's find the velocity of the car at the end of the initial acceleration phase.

Using the equation of motion:

v = u + at,

where:
v = final velocity
u = initial velocity (which is 0 m/s since the car starts from rest)
a = acceleration (1.6 m/s^2)
t = time (10.0 s)

Plugging in the values:

v = 0 + (1.6 m/s^2) * (10.0 s)
v = 16.0 m/s

At the end of the acceleration phase, the car's velocity is 16.0 m/s.

Next, let's find the distance traveled during the acceleration phase using the equation:

s = ut + 0.5 * a * t^2,

where:
s = distance
u = initial velocity (0 m/s)
t = time (10.0 s)
a = acceleration (1.6 m/s^2)

Plugging in the values:

s = 0 + 0.5 * (1.6 m/s^2) * (10.0 s)^2
s = 0 + 0.5 * (1.6 m/s^2) * 100.0 s^2
s = 0 + 0.5 * 16.0 m
s = 0 + 8.0 m
s = 8.0 m

The car has traveled a distance of 8.0 meters during the acceleration phase.

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

We need to find the final velocity and distance traveled during this phase.

Using the same equation of motion as before, but now with a negative acceleration:

v = u + at,

where:
v = final velocity
u = initial velocity (which is 16.0 m/s from the previous calculation)
a = acceleration (-3.0 m/s^2, since it is negative)
t = time (1.0 s)

Plugging in the values:

v = 16.0 m/s + (-3.0 m/s^2) * (1.0 s)
v = 16.0 m/s - 3.0 m/s
v = 13.0 m/s

At the end of the deceleration phase, the car's velocity is 13.0 m/s (in the opposite direction of its initial motion).

To find the distance traveled during the deceleration phase, we'll use the same equation as before:

s = ut + 0.5 * a * t^2,

where:
s = distance
u = initial velocity (16.0 m/s from the previous calculation)
t = time (1.0 s)
a = acceleration (-3.0 m/s^2)

Plugging in the values:

s = 16.0 m/s * (1.0 s) + 0.5 * (-3.0 m/s^2) * (1.0 s)^2
s = 16.0 m + 0.5 * (-3.0 m/s^2) * (1.0 s)^2
s = 16.0 m - 0.5 * 3.0 m
s = 16.0 m - 1.5 m
s = 14.5 m

The car has traveled a distance of 14.5 meters during the deceleration phase.

Therefore, at the end of the braking period, the car's velocity is -13.0 m/s and it has traveled a total distance of 8.0 m + 14.5 m = 22.5 meters.