A car starts from rest and travels for 6.0 s with a uniform acceleration of +2.4 m/s2. The driver then applies the brakes, causing a uniform acceleration of −1.0 m/s2. If the brakes are applied for 3.0 s, 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?

4.65

To solve this problem, we can use the equations of motion to calculate the final velocity and total distance traveled by the car.

(a) How fast is the car going at the end of the braking period?

We'll first calculate the final velocity during the acceleration phase and then use that to calculate the final velocity during the deceleration phase.

For the acceleration phase:
Initial velocity (u) = 0 m/s
Acceleration (a) = +2.4 m/s^2
Time (t) = 6.0 s

Using the equation:
v = u + at
where:
v = final velocity

Substituting the given values:
v = 0 + (2.4)(6.0)
v = 14.4 m/s

Now, for the deceleration phase:
Initial velocity (u) = 14.4 m/s (as this is the final velocity from the acceleration phase)
Acceleration (a) = -1.0 m/s^2 (negative because it's deceleration)
Time (t) = 3.0 s

Using the same equation:
v = u + at

Substituting the given values:
v = 14.4 + (-1.0)(3.0)
v = 14.4 - 3.0
v = 11.4 m/s

Therefore, the car is going at a speed of 11.4 m/s at the end of the braking period.

(b) How far has the car gone?

To calculate the distance traveled during each phase, we'll use the equation:

s = ut + (1/2)at^2
where:
s = distance traveled

For the acceleration phase:
Initial velocity (u) = 0 m/s
Acceleration (a) = +2.4 m/s^2
Time (t) = 6.0 s

Substituting the given values:
s = (0)(6.0) + (1/2)(2.4)(6.0)^2
s = 0 + (1/2)(2.4)(36)
s = 0 + 43.2
s = 43.2 m

For the deceleration phase:
Initial velocity (u) = 14.4 m/s
Acceleration (a) = -1.0 m/s^2
Time (t) = 3.0 s

Substituting the given values:
s = (14.4)(3.0) + (1/2)(-1.0)(3.0)^2
s = 43.2 + (-1.5)
s = 41.7 m

Therefore, the car has traveled a total distance of 43.2 m (during the acceleration phase) + 41.7 m (during the deceleration phase) = 84.9 m.

To determine the answers to (a) and (b), we need to use the equations of motion.

First, let's calculate the velocity of the car after the initial acceleration period:

(a) How fast is the car going at the end of the braking period?

We can use the equation:

v = u + at

where v is the final velocity, u is the initial velocity (which is 0 since the car starts from rest), a is the acceleration, and t is the time.

Given:
- Initial velocity (u) = 0 m/s
- Acceleration (a) during acceleration period = +2.4 m/s²
- Time (t) during acceleration period = 6.0 s

Plugging in these values, the equation becomes:

v = 0 + 2.4 * 6.0
v = 14.4 m/s

Therefore, at the end of the acceleration period, the car is traveling at a velocity of 14.4 m/s.

Now, let's calculate the distance the car has traveled during the entire motion:

(b) How far has the car gone?

We can use the equation:

s = ut + (1/2) * a * t^2

where s is the distance traveled.

For the first part of the motion (acceleration):
- Initial velocity (u) = 0 m/s
- Acceleration (a) = +2.4 m/s²
- Time (t) = 6.0 s

Plugging in these values, the equation becomes:

s1 = 0 * 6 + (1/2) * 2.4 * (6.0)^2
s1 = 0 + 7.2 * 36
s1 = 259.2 m

The car has traveled a distance of 259.2 m during the acceleration period.

For the second part of the motion (braking):
- Initial velocity (u) = 14.4 m/s (the final velocity from the previous calculation)
- Acceleration (a) = -1.0 m/s²
- Time (t) = 3.0 s

Plugging in these values, the equation becomes:

s2 = 14.4 * 3 + (1/2) * (-1.0) * (3.0)^2
s2 = 43.2 - 4.5
s2 = 38.7 m

The car has traveled a distance of 38.7 m during the braking period.

To find the total distance traveled, we sum the distances from the acceleration and braking periods:

Total distance (s) = s1 + s2
Total distance = 259.2 + 38.7
Total distance = 297.9 m

Therefore, the car has traveled a total distance of 297.9 m during the entire motion.

(a) After 6.0 s, the car's velocity is

a*t = 2.4 * 6 = 14.4 m/s and it will have travelled
X = (a/2) t^2 = 86.4 m.
Three seconds of braking at -1.0 m/s^2 then reduces the velocity, by 3.0 m/s, from
14.4 to 11.4 m/s.
(b) Additional distance travelled while braking
= (avg. speed durng interval)*3.0 s
= 12.9 * 3 = 38.7 m
Add 38.7 m to 86.4 m for the total distance travelled during the 9.0 seconds.