A bus accelerates at 2.4 m/s2 from rest for 10 s. It then travels at constant speed for 25 s, after which it slows to a stop with an acceleration of magnitude 1.5 m/s2.

(a) What is the total distance that the bus travels?
(b) What was its average velocity?
If possible, could you explain things and how you got the formulas too? Thank you!

V1 = a*T1 = 2.4 * 10 = 24 m/s.

d1 = 0.5a*T1^2 = 1.2*10^2 = 120 m.
a. d2 = V1*T2 = 24 * 25 = 600 m.

V^2 = V1^2 + 2a*d3 = 0
d3 = -V1^2/2a = -(24^2)/-3 = 192 m.

d = d1+d2+d3 = 120 + 600 + 192 = 912 m. = Total distance traveled.

b. V = V1 + a*T3 = 0.
T3 = -V1/a = -24/-1.5 = 16 s.

T = T1+T2+T3 = 10 + 25 + 16 = 51 s.

V = d/T = 912m/51s. = 17.9 m/s. = Avg.
velocity.

To solve this problem, we'll break it down into three sections: the initial acceleration, the constant speed, and the final deceleration.

Section 1: Initial acceleration
We are given the acceleration (a) as 2.4 m/s^2 and the time (t) as 10 s. Since the bus starts from rest, we can use the formula:
distance = (initial velocity * time) + (0.5 * acceleration * time^2)

Since the initial velocity is 0, we can simplify the formula to:
distance = 0.5 * acceleration * time^2

Substituting in the given values:
distance1 = 0.5 * 2.4 * (10)^2 = 120 m

Section 2: Constant speed
The bus travels at a constant speed for 25 s. Since the speed is constant, there is no acceleration and the formula used is:
distance = speed * time

We are not given the speed, but we know the time and distance traveled in the previous section, so we can find the speed:
speed = distance1 / time = 120 m / 10 s = 12 m/s

Now, using the speed and time for this section:
distance2 = speed * time = 12 m/s * 25 s = 300 m

Section 3: Final deceleration
We are given the deceleration (a) as -1.5 m/s^2 (negative since it's slowing down) and we need to find the distance traveled until the bus stops. Since it starts at a known speed (12 m/s), we can use the formula:
final velocity^2 = initial velocity^2 + 2 * acceleration * distance

Since the final velocity is 0 (since the bus stops), we can simplify the formula to:
distance = (final velocity^2) / (2 * acceleration)

Substituting in the given values:
distance3 = (0^2) / (2 * -1.5) = 0 m

(a) Total distance traveled:
The total distance traveled is the sum of the distances from each section:
total distance = distance1 + distance2 + distance3 = 120 m + 300 m + 0 m = 420 m

(b) Average velocity:
Average velocity is defined as total distance traveled divided by total time taken. The total time taken is equal to the sum of the times for each section, which is 10 s + 25 s = 35 s.
average velocity = total distance / total time = 420 m / 35 s = 12 m/s

So, the total distance traveled by the bus is 420 meters, and its average velocity is 12 m/s.