(455) Uniformly accelerated motion 2 [IB Physics SL/HL]

Uniformly Accelerated Motion Example

• Situation: Person pushed off a 325 m high cliff.

• Goal: Determine the speed upon hitting the ground.

Key Concepts

• Initial Speed (U): Implied to be 0 m/s (not running).

• Final Speed (V): Unknown, we are solving for this.

• Acceleration (a): Due to gravity, which is -9.81 m/s² (negative as it acts downward).

• Displacement (s): 325 m downward.

Motion Equations

• Relevant Equation: V² = U² + 2as

  • This equation does not involve time (T), simplifying calculation.

Application of the Equation

• Substituting Known Values:

  • U = 0 m/s

  • a = -9.81 m/s²

  • s = -325 m

  • Thus, V² = 0 + 2(-9.81)(-325)

Calculation Steps

• Calculation:

  • V² = 2 * 9.81 * 325

  • V = ±√(2 * 9.81 * 325)

  • Calculate to find V = ±79.85 m/s

Result Interpretation

• Final Speed: V ≈ 79.9 m/s (or about 80 m/s).

• Direction: Downward, hence technically velocity would be -79.9 m/s.

• Conversion to km/h: Approximately 280 km/h, illustrating a high speed.

Conclusion

• Understanding uniformly accelerated motion with gravity leads to determining speeds and interpreting results with proper sign conventions.