Answer :
a) It takes you 1.893s to reach the opened door and jump in.
b) The maximum time that you can wait before starting to run and still catch the bus is 1.27s
The time which the passenger can delay and still catch the bus must be less than this.
From the question:
Distance between the passenger and the door of the bus d = 9
Acceleration of the bus a¹ = 1.0m/s²
Speed of the passenger v = 5.7m/s
Time taken for you to catch up with the bus must occur at the exact position with the bus,
Let the position = y
The time taken for you to reach the point:
t = total speed/distance
t = y + 9/5.7 ..................... eqn 1
The time taken for the bus to reach the same point;
y = vt + 1/2at²
Initial velocity of the bus is zero
y = 0.5*(1)t²
y = t²/2
t = √2y .......................eqn 2
We solve eqn (1) and (2) together:
9 + y = 5.7 * √2y
81 + y² + 18y = 64.98y
y² - 46.98y + 81 = 0
y = 45.187m or 1.792m
Now we take the smallest distance: x=1.792 m
Time taken for both to reach this point: (put x in eq. (1))
t = 1.792+9/5.7
t = 1.893s
For the maximum waiting time we find have:
Relative acceleration, ar = -1m/s⁻²since the bus is moving away from the intended
The final relative velocity for the passenger to catch it, vr = 0m/s⁻¹
Using equation of motion:
s = 9 + 1/2 * (-1)t²
also the initial relative velocity:
vr = ur + ar.t
ur = 5.7 - t
and
vr² = ur² + 2ar.s
0² = (5.7 - 1)² - 2 * 1 * (9 - 0.5t²)
t = 1.27s
Here's the complete question:
You are 9.0 m from the door of your bus, behind the bus, when it pulls away with an acceleration of 1.0 m/s2. You instantly start running toward the still-open door at 5.7 m/s.
How long does it take for you to reach the open door and jump in?
What is the maximum time you can wait before starting to run and still catch the bus?
Learn more about calculating maximum time from:
https://brainly.com/question/14698468?referrer=searchResults
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