Motion in a Straight Line - Lecture 05

Release of Object from Moving Platform

  • When a ball is released from a moving train, its initial velocity is the same as the train's velocity at the moment of release. This is because the ball is already in motion with the train before it is released. Inertia ensures it maintains this initial velocity.

  • However, the acceleration of the ball changes once it's in the air; it is then only influenced by gravity, acting downwards. Air resistance is often ignored for simplicity, but in real-world scenarios, it would also affect the ball's motion.

Variable Acceleration Motion

  • Velocity is defined as the rate of change of position with respect to time: V = \frac{d\vec{r}}{dt}

  • In one dimension (along the x-axis), velocity is V = \frac{dx}{dt}. Here, x represents the position of the object along the x-axis, and t represents time. The derivative \frac{dx}{dt} gives the instantaneous velocity at any given moment.

  • Speed is the magnitude of the velocity: V = ||\frac{d\vec{r}}{dt}||. Speed is always a non-negative scalar value, representing how fast an object is moving, regardless of direction.

  • Acceleration is defined as the rate of change of velocity with respect to time: a = \frac{dv}{dt}. It indicates how quickly the velocity of an object is changing. Acceleration can be positive (speeding up), negative (slowing down), or zero (constant velocity).

One-Dimensional Motion

  • a = \frac{dv}{dt}, V = \frac{dx}{dt}

  • a = v \frac{dv}{dx}

Position Vector

  • \vec{V} represents velocity. Velocity is a vector quantity, having both magnitude (speed) and direction.

  • \vec{a} represents acceleration. Acceleration is also a vector quantity, indicating the rate of change of velocity, including changes in direction.

  • t represents time. Time is a scalar quantity, representing the duration of an event.

  • S represents displacement. Displacement is a vector quantity that refers to the change in position of an object. It is the shortest distance between the initial and final positions and has a direction.

Differentiation and Integration in Kinematics

  • Given position x, we differentiate to find velocity v and differentiate velocity to find acceleration a. Differentiation provides the instantaneous rate of change.

  • Conversely, we integrate acceleration to find velocity and integrate velocity to find displacement. Integration is the reverse process of differentiation and allows us to find the cumulative effect over time.

Examples

  • Example 1: If x = 2t^3, then v = \frac{dx}{dt} = 6t^2 and a = \frac{dv}{dt} = 12t. In this example, the position x is given as a function of time. By differentiating once, we find the velocity v, and by differentiating again, we find the acceleration a.

Integrating Acceleration to Find Velocity

  • a = \frac{dv}{dt} \implies \int dv = \int a dt

  • If a = 2t^2, then \int dv = \int 2t^2 dt. Here, we integrate both sides of the equation with respect to time to find the velocity as a function of time.

Integrating Acceleration to Find Velocity with Position

  • a = 2x^2 = v \frac{dv}{dx} \implies \int v dv = \int 2x^2 dx

General Integration

  • \int dx = f(t)

  • \int 1 dx

Variable Acceleration Motion Problems

  • Given: A particle is moving on the x-axis, and its position varies with time as x = t^2 - 10t + 15 meters.

    • Find:

    • Velocity and acceleration at time t.

    • Initial position and initial velocity.

    • Position when velocity becomes zero.

    • Displacement and distance traveled in the first 9 seconds.

Solutions

  1. Velocity and acceleration:

    • v = \frac{dx}{dt} = 2t - 10 \text{ m/s}

    • a = \frac{dv}{dt} = 2 \text{ m/s}^2

  2. Initial position and velocity (at t=0):

    • x_i = 0^2 - 10(0) + 15 = 15 \text{ m}

    • v_i = 2(0) - 10 = -10 \text{ m/s}

  3. Position when velocity is zero:

    • 2t - 10 = 0 \implies t = 5 \text{ s}

    • x(t=5) = (5)^2 - 10(5) + 15 = -10 \text{ m}

  4. Displacement and distance in the first 9 seconds:

    • x(t=9) = (9)^2 - 10(9) + 15 = 6 \text{ m}

    • Displacement: S = x(t=9) - x(t=0) = 6 - 15 = -9 \text{ m}

Additional Problems

  1. Ball Projected from a Tower

    • A ball is projected upwards with a speed of 20 m/s from the top of a 160 m tower.

    • Find:

      • Time of flight (T).

      • Speed with which the ball hits the ground.

      • Maximum height (H) reached by the ball.

      • Total distance traveled by the ball.

    • Solution:

      • Using S = ut + \frac{1}{2}at^2: -160 = 20t + \frac{1}{2}(-10)t^2

      • Simplified: t^2 - 4t - 32 = 0

      • Solving for t: t = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-32)}}{2(1)} = \frac{4 \pm \sqrt{16 + 128}}{2} = \frac{4 \pm 12}{2}

      • Since time cannot be negative, t = \frac{4 + 12}{2} = 8 \text{ s}

      • Final velocity: V = u + at = 20 + (-10)(8) = -60 \text{ m/s}

  1. Ball Released from a Tower

    • A ball is released from the top of a tower and travels 100 m in the last 2 seconds.

    • Find the height of the tower.

    • Solution:

      • Using S = ut + \frac{1}{2}at^2 (since u = 0): H = \frac{1}{2}gt^2 = 5t^2

      • H - 100 = 5(t-2)^2

  1. Particle Projected Upwards with Air Resistance

    • A particle is projected upwards with speed u from the ground.

    • Air resistance produces a constant retardation of a (where a < g).

    • Find:

      • Maximum height (H).

      • Time of ascent (t_1).

      • Time of descent (t_2).

      • Total time of flight (T).

      • Speed with which the ball hits the