Lecture 8 1d Kinematics

Velocity-Time Diagrams and Vertical Motion Under Gravity

1. Example: Car Motion

• A car starts from rest with an acceleration of 1 m/s².

• Maximum velocity reached: 10 m/s.

• Stages of motion broken down:

  • 1st Stage:

    • From the diagram, with slope a = 1 (indicating acceleration),

      • Formula: a = (v - u) / t gives:

      • t₁ = (v - u) / a = (10 m/s - 0) / 1 m/s² = 10 s.

    • Area under the velocity-time graph (distance):

      • Formula: s₁ = 1/2 * base * height,

      • s₁ = 1/2 * t₁ * v = 1/2 * 10 s * 10 m/s = 50 m.

  • 2nd Stage:

    • Covers a distance of 100 m at 10 m/s:

      • Formula: Distance = speed × time,

      • 100 m = 10 m/s × t₂ → t₂ = 10 s.

• Total time taken: t₁ + t₂ = 10 s + 10 s = 20 s.

• Total distance covered: s₁ + 100 m = 50 m + 100 m = 150 m.

2. Vertical Motion Under Gravity

• Galileo's Principle: All objects fall under gravity with the same acceleration regardless of mass.

  • Acceleration due to gravity (g): g = 9.81 m/s², approximated as 10 m/s² near Earth's surface.

3. Example: Vertical Throw

• A ball is thrown vertically upwards with an initial speed (u) of 30 m/s.

• To calculate maximum height (h):

  • At maximum height, the final velocity (v) is 0:

    • v = u + at → 0 = 30 m/s - g * t.

  • Solving for time (t) when v = 0:

    • t = 30/g ≈ 3 s.

  • Height (s) formula:

    • s = ut + (1/2)at² = 30t - 5t² (approx).

  • At t = 3 s:

    • h = 30(3) - 5(3)² = 90 m - 45 m = 45 m.

• Alternative Calculation:

  • Using v² - u² = 2as:

    • 0 - (30)² = 2(-g)h,

    • h = (30²)/(2 * 10) = 450/20 = 45 m.

4. Example: Dropped Mass

• A mass is dropped from height h:

  • Determine time to reach ground and final velocity:

    • Initial conditions: at t=0, s = h and u = 0:

      • v = u + at → v = -gt (negative due to downward direction).

    • Integrate to find how long it takes to reach the ground:

      • Formula: s(t) = s(0) + ∫v dt = h - (1/2)gt² (integration from 0 to T).

    • Setting s at t=T to 0:

      • h = (1/2)gT² → T = √(2h/g).

• Final velocity at impact:

  • v = -gT = -g√(2h/g) = -√(2gh).

5. Dimensional Analysis

• Check dimensional consistency of T = √(2h/g):

  • Dimensions of left side: [T] = Time.

  • Dimensions of right side: √(s²h g) gives:

    • s | Length, | s | Length, | (s| Length)/(T| Time)² | → This confirms both sides match in dimensions.

• Conclusion: A formula such as T = √(gh) is impossible as it results in mismatched dimensions.