A bus travels 400 m between two stops. It starts from rest and accelerates at 1.50 m/s^2 until it reaches a velocity of 9.00 m/s. The bus continues at this velocity and then decelerates at 2 m/s^2 until it comes to a halt. Find the total time required for the journey.
i rechecked it and i got 44.33 something????
Well, this bus seems to have quite the need for speed! Let's calculate the time it takes for each segment of the journey.
First, let's find the time it takes for the bus to reach a velocity of 9.00 m/s. We can use the formula:
v = u + at,
where v is the final velocity, u is the initial velocity (0 m/s, as the bus starts from rest), a is the acceleration (1.50 m/s^2), and t is the time.
Using this formula, we have:
9.00 = 0 + (1.50)t.
Simplifying, we find:
t = 9.00 / 1.50 = 6 seconds.
So, it takes 6 seconds for the bus to reach a velocity of 9.00 m/s.
Now, to calculate the time it takes for the bus to decelerate and come to a halt, we can use the equation:
v^2 = u^2 + 2as,
where v is the final velocity (0 m/s, as the bus comes to a halt), u is the initial velocity (9.00 m/s), a is the acceleration (-2 m/s^2, since it's decelerating), and s is the distance (400 m).
Plugging in the values, we get:
0^2 = 9.00^2 + 2(-2)s.
Simplifying, we find:
s = (81.00) / 4 = 20.25 meters.
Now we have the distance and the acceleration for the decelerating phase. Using the formula:
s = ut + 0.5at^2,
where s is the distance (20.25 m), u is the initial velocity (9.00 m/s), a is the acceleration (-2 m/s^2), and t is the time, we can solve for t.
Plugging in the values, we have:
20.25 = 9.00t + 0.5(-2)t^2.
Simplifying, we get a quadratic equation:
-1t^2 + 9t + 20.25 = 0.
Solving this quadratic equation, we find:
t = 2.25 seconds.
So, it takes 2.25 seconds for the bus to decelerate and come to a halt.
Now, to find the total time required for the journey, we simply add up the times for each segment:
Total time = time to accelerate + time to decelerate
Total time = 6 + 2.25 = 8.25 seconds.
Voila! The total time required for the journey is 8.25 seconds. Keep in mind, though, that this calculation doesn't account for any stops, traffic, or clown cars blocking the way!
To find the total time required for the journey, we need to consider the time taken during acceleration, the time taken while traveling at a constant velocity, and the time taken during deceleration.
First, let's calculate the time taken during acceleration. We can use the equation of motion:
v = u + at
where
v = final velocity (9.00 m/s)
u = initial velocity (0 m/s, as the bus starts from rest)
a = acceleration (1.50 m/s^2)
t = time taken during acceleration (unknown)
Rearranging the equation to solve for time (t), we have:
t = (v - u) / a
Substituting the given values, we get:
t = (9.00 m/s - 0 m/s) / 1.50 m/s^2
t = 6.00 s
Therefore, the time taken during acceleration is 6.00 seconds.
Next, let's calculate the time taken at a constant velocity. We can use the formula:
d = vt
where
d = distance traveled at a constant velocity (400 m)
v = velocity during that time (9.00 m/s)
t = time taken at constant velocity (unknown)
Rearranging the formula to solve for time (t), we have:
t = d / v
Substituting the given values, we get:
t = 400 m / 9.00 m/s
t = 44.44 s (rounded to two decimal places)
Therefore, the time taken at the constant velocity is approximately 44.44 seconds.
Finally, let's calculate the time taken during deceleration using the same equation of motion:
t = (v - u) / a
where
v = final velocity (0 m/s, as the bus comes to a halt)
u = initial velocity (9.00 m/s, since the velocity is constant until deceleration)
a = deceleration (-2 m/s^2, as it is in the opposite direction of acceleration)
Substituting the given values, we get:
t = (0 m/s - 9.00 m/s) / -2 m/s^2
t = 4.5 s
Therefore, the time taken during deceleration is 4.5 seconds.
To find the total time, we add the times taken during acceleration, constant velocity, and deceleration:
Total time = time taken during acceleration + time taken at constant velocity + time taken during deceleration
Total time = 6.00 s + 44.44 s + 4.5 s
Total time = 54.94 s
Therefore, the total time required for the journey is approximately 54.94 seconds.
initially, v = at, so t = 9.00/1.50 = 6s
The distance covered is 1/2 at^2 = 27m
When decelerating, it takes 9/2 = 4.5s to stop
In that time, it covered 9*4.5 - 1/2 * 2 * 4.5^2 = 20.25m
So the distance covered at 9 m/s is 400-27-0.25 = 352.75
The total time is thus 6 + 352.75/9 + 4.5 = 49.69 s