A car starts from rest and travels for 4.93 s with a uniform acceleration of +1.53 m/s2. The driver then applies the brakes, causing a uniform acceleration of -1.99 m/s2. If the brakes are applied for 2.83 s, how fast is the car going at the end of the braking period?

To find the final velocity of the car at the end of the braking period, we need to break down the problem into two parts: the initial acceleration phase and the braking phase.

1. Initial acceleration phase:
During the initial acceleration phase, we can use the kinematic equation:

v = u + at

Where:
- v is the final velocity
- u is the initial velocity (which is 0 m/s since the car starts from rest)
- a is the acceleration during this phase (+1.53 m/s^2)
- t is the time during this phase (4.93 s)

Using the given values, we can calculate the final velocity at the end of the initial acceleration phase:

v1 = 0 + (1.53 m/s^2) * (4.93 s)
v1 = 7.5569 m/s

2. Braking phase:
During the braking phase, the acceleration is negative (-1.99 m/s^2), and the time is given as 2.83 seconds. Again, we can use the same kinematic equation to find the final velocity:

v2 = v1 + (-a) * t

Where:
- v2 is the final velocity at the end of the braking phase
- v1 is the velocity at the end of the initial acceleration phase (7.5569 m/s)
- a is the acceleration during the braking phase (-1.99 m/s^2)
- t is the time during the braking phase (2.83 s)

Plugging in the values, we can calculate the final velocity at the end of the braking phase:

v2 = 7.5569 m/s + (-1.99 m/s^2) * (2.83 s)
v2 = 2.1285 m/s

Therefore, the car is traveling at a speed of 2.1285 m/s at the end of the braking period.