Math Problem Statement

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/s/s. You instantly start running toward the still-open door at 5.5 m/s. What is the maximum time you can wait before starting to run and still catch the bus?

Solution

To solve this problem, we need to determine the maximum time you can wait before starting to run and still catch the bus.

Step 1: Define Variables

  • Distance between you and the bus initially: d0=9.0md_0 = 9.0 \, \text{m}
  • Your running speed: v=5.5m/sv = 5.5 \, \text{m/s}
  • Acceleration of the bus: a=1.0m/s2a = 1.0 \, \text{m/s}^2
  • Time you wait before starting to run: twaitt_{\text{wait}}

Step 2: Set Up Equations

  • The bus starts from rest with acceleration, so its position after tt seconds is given by: xbus(t)=12at2=12(1.0)t2=0.5t2x_{\text{bus}}(t) = \frac{1}{2} a t^2 = \frac{1}{2} (1.0) t^2 = 0.5 t^2

  • You start running after a delay of twaitt_{\text{wait}}. The time you spend running is ttwaitt - t_{\text{wait}}, and since you're moving at constant speed, your position is: xyou(t)=v(ttwait)=5.5(ttwait)x_{\text{you}}(t) = v (t - t_{\text{wait}}) = 5.5 (t - t_{\text{wait}})

Initially, you're 9.0 meters behind the bus, so the relative position between you and the bus must satisfy: xyou(t)=xbus(t)+d0x_{\text{you}}(t) = x_{\text{bus}}(t) + d_0 Substituting the expressions for xyou(t)x_{\text{you}}(t) and xbus(t)x_{\text{bus}}(t): 5.5(ttwait)=0.5t2+9.05.5 (t - t_{\text{wait}}) = 0.5 t^2 + 9.0

Step 3: Solve for twaitt_{\text{wait}}

We will now solve this equation for the maximum twaitt_{\text{wait}} that ensures you catch the bus.

First, expand and rearrange the equation: 5.5t5.5twait=0.5t2+9.05.5 t - 5.5 t_{\text{wait}} = 0.5 t^2 + 9.0 0.5t25.5t+9.0=5.5twait0.5 t^2 - 5.5 t + 9.0 = -5.5 t_{\text{wait}} Now, express twaitt_{\text{wait}}: twait=0.5t25.5t+9.05.5t_{\text{wait}} = \frac{0.5 t^2 - 5.5 t + 9.0}{-5.5} To find the maximum twaitt_{\text{wait}}, we will need to determine the time tt at which the equation holds true by solving for when you and the bus meet.

I'll calculate this now.The times when you and the bus could meet are t=2.0secondst = 2.0 \, \text{seconds} and t=9.0secondst = 9.0 \, \text{seconds}. The relevant time for catching the bus is t=9.0secondst = 9.0 \, \text{seconds}, since t=2.0secondst = 2.0 \, \text{seconds} represents a scenario where you wouldn't catch up yet.

Step 4: Maximum Time Before Running

Using the time t=9.0secondst = 9.0 \, \text{seconds}, we can now calculate the maximum time you can wait before running, which is twaitt_{\text{wait}}. The equation for twaitt_{\text{wait}} was: twait=0.5t25.5t+9.05.5t_{\text{wait}} = \frac{0.5 t^2 - 5.5 t + 9.0}{-5.5} Substitute t=9.0secondst = 9.0 \, \text{seconds} into the equation.

I'll compute this now.It seems like I can’t do more advanced data analysis right now. Please try again later.

In the meantime, if you'd like help understanding the final steps of the problem or want to try manually calculating twaitt_{\text{wait}}, feel free to ask! Here's a reminder of the last equation you would need:

twait=0.5×925.5×9+9.05.5t_{\text{wait}} = \frac{0.5 \times 9^2 - 5.5 \times 9 + 9.0}{-5.5}

Let me know if you'd like further assistance with it or if you have any questions.

5 Related Questions:

  1. What happens if your running speed increases?
  2. How would the answer change if the bus had a different acceleration?
  3. What if the distance to the bus was greater than 9.0 m?
  4. How would air resistance affect the time to catch the bus?
  5. What if you started running before the bus began accelerating?

Tip:

Always keep track of units during calculations to ensure accuracy, especially in problems involving multiple physical quantities like speed and acceleration.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Quadratic Equations
Relative Motion

Formulas

Position of the bus: x_bus(t) = 0.5 * a * t^2
Position of the runner: x_you(t) = v * (t - t_wait)
Relative position equation: x_you(t) = x_bus(t) + d_0

Theorems

Kinematic equations of motion
Quadratic formula

Suitable Grade Level

Grades 10-12