Math Problem Statement

Kate Ashley was driving her car filled with drugs at the trunk at 100km/h when she saw a police car. She accelerated to 180km/h in just 1 second to avoid the police. What is the acceleration?

Solution

To calculate the acceleration, we use the formula:

a=ΔvΔta = \frac{{\Delta v}}{{\Delta t}}

Where:

  • aa is the acceleration,
  • Δv\Delta v is the change in velocity,
  • Δt\Delta t is the change in time.

In this case:

  • The initial velocity vi=100km/hv_i = 100 \, \text{km/h},
  • The final velocity vf=180km/hv_f = 180 \, \text{km/h},
  • The time taken is 1 second.

First, convert the velocities from km/h to m/s. We know that: 1km/h=10003600m/s=0.2778m/s1 \, \text{km/h} = \frac{1000}{3600} \, \text{m/s} = 0.2778 \, \text{m/s}

Thus: vi=100km/h=100×0.2778m/s=27.78m/sv_i = 100 \, \text{km/h} = 100 \times 0.2778 \, \text{m/s} = 27.78 \, \text{m/s} vf=180km/h=180×0.2778m/s=50m/sv_f = 180 \, \text{km/h} = 180 \times 0.2778 \, \text{m/s} = 50 \, \text{m/s}

Now, calculate the acceleration: a=vfviΔt=50m/s27.78m/s1s=22.22m/s2a = \frac{{v_f - v_i}}{{\Delta t}} = \frac{{50 \, \text{m/s} - 27.78 \, \text{m/s}}}{1 \, \text{s}} = 22.22 \, \text{m/s}^2

Thus, the acceleration is 22.22 m/s².

Would you like further details on how acceleration works, or any other questions?


Here are 5 related questions:

  1. What would be the stopping distance if Kate Ashley decelerated suddenly?
  2. How much time would it take for her to reach 250 km/h at the same acceleration?
  3. How does the acceleration compare with typical car acceleration rates?
  4. How does acceleration affect fuel consumption in cars?
  5. How could the speed limit laws affect this scenario in real life?

Tip: Always ensure to convert all units to the same system (e.g., m/s, s) before performing calculations to avoid errors.

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Math Problem Analysis

Mathematical Concepts

Kinematics
Acceleration

Formulas

a = (vf - vi) / t
1 km/h = 0.2778 m/s

Theorems

Newton's Second Law of Motion

Suitable Grade Level

Grades 9-12