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Motion along a straight line Simplified Revision Notes

Revision notes with simplified explanations to understand Motion along a straight line quickly and effectively.

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4.1.3 Motion along a straight line

  1. Speed: A scalar quantity that describes how fast an object is moving, without any regard for direction.
  2. Displacement (ss): A vector quantity representing the overall distance moved from the initial position, including direction.
  3. Velocity (vv): The rate of change of displacement with respect to time, given by:
v=ΔsΔtv = \frac{\Delta s}{\Delta t}
  1. Acceleration (aa): The rate of change of velocity with respect to time, given by:
a=ΔvΔta = \frac{\Delta v}{\Delta t}

Types of Velocity

  • Instantaneous Velocity: The velocity at a specific moment in time. It can be found by determining the gradient of a tangent line on a displacement-time graph.
  • Average Velocity: The velocity over a specified time period. It is calculated by dividing the total displacement by the total time taken.

Uniform Acceleration

When an object experiences constant acceleration, we describe it as having uniform acceleration.

image

Interpreting Motion Graphs

  1. Acceleration-Time Graphs:
  • Show how velocity changes over time.
  • The area under the graph gives the change in velocity.
  1. Velocity-Time Graphs:
  • Show how velocity varies over time.
  • The gradient of this graph represents acceleration.
  • The area under the graph represents displacement.
  1. Displacement-Time Graphs:
  • Show the change in displacement over time.
  • The gradient of this graph represents velocity.
image

Equations of Motion for Uniform Acceleration

For objects moving with uniform acceleration, the following equations of motion apply:

  1. v=u+atv = u + at
  2. s=(u+v2)ts = \left(\frac{u + v}{2}\right)t
  3. s=ut+12at2s = ut + \frac{1}{2}at^2
  4. v2=u2+2asv^2 = u^2 + 2as

Where:

  • ss = displacement
  • uu = initial velocity
  • vv = final velocity
  • aa = acceleration
  • tt = time Tip: When solving problems, list out known and unknown values and choose the equation that best fits.
infoNote

Worked Example Problem: A stone is dropped from a bridge that is 5050 m above the water. Calculate:

  • The final velocity vv of the stone just before it hits the water.
  • The time tt it takes to reach the water. Solution:
  1. List known values:
  • s=50ms = 50 \, \text{m}
  • u=0m/su = 0 \, \text{m/s} (since it's dropped)
  • a=9.81m/s2a = 9.81 \, \text{m/s}^2 (acceleration due to gravity)
  • v=?,t=? v = ? , t = ?
  1. Calculate Final Velocity: Using the equation v2=u2+2asv^2 = u^2 + 2as :
v2=02+2×9.81×50v^2 = 0^2 + 2 \times 9.81 \times 50v2=981v^2 = 981 v=981=:success[31.3m/s]v = \sqrt{981} = :success[31.3 \, \text{m/s}]
  1. Calculate Time: Using the equation s=ut+12at2s = ut + \frac{1}{2}at^2 :
50=0×t+12×9.81×t250 = 0 \times t + \frac{1}{2} \times 9.81 \times t^2 50=4.905t250 = 4.905t^2 t2=504.905=10.19t^2 = \frac{50}{4.905} = 10.19 t=10.19=:success[3.19s]t = \sqrt{10.19} = :success[3.19 \, \text{s}]
infoNote

Key Points

  1. Speed and Velocity: Speed is scalar, while velocity is a vector.
  2. Acceleration: Represents change in velocity over time.
  3. Uniform Acceleration Equations: Use these equations to solve motion problems with constant acceleration.
  4. Graph Analysis: Displacement-time, velocity-time, and acceleration-time graphs each provide unique insights into motion.
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