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Physics: Motion and Mechanics - Equations of Motion

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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SUVAT Variables: Linear motion with constant acceleration is described using five key variables: ss (displacement), uu (initial velocity), vv (final velocity), aa (acceleration), and tt (time).

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Uniform Acceleration: The equations of motion only apply when the acceleration aa is constant (uniform). If acceleration changes, these specific formulae cannot be used directly.

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Scalar vs. Vector: Displacement (ss), velocity (vv), and acceleration (aa) are vectors, meaning they have both magnitude and direction. Distance and speed are their scalar counterparts.

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Directional Sign Convention: It is crucial to define a positive direction (e.g., right or up is positive). If an object slows down, its acceleration is usually represented as a negative value (deceleration), such as a=−2 m/s2a = -2 \text{ m/s}^2.

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Starting from Rest: If a problem states an object starts from rest, the initial velocity is u=0 m/su = 0 \text{ m/s}. If it comes to a stop, the final velocity is v=0 m/sv = 0 \text{ m/s}.

📐Formulae

v=u+atv = u + at

s=ut+12at2s = ut + \frac{1}{2}at^2

v2=u2+2asv^2 = u^2 + 2as

s=(u+v)2×ts = \frac{(u + v)}{2} \times t

💡Examples

Problem 1:

An athlete starts from rest and accelerates at a constant rate of 2 m/s22 \text{ m/s}^2 for 5 s5 \text{ s}. Calculate the displacement covered by the athlete.

Solution:

Given: u=0 m/su = 0 \text{ m/s}, a=2 m/s2a = 2 \text{ m/s}^2, t=5 st = 5 \text{ s}. We use the formula: s=ut+12at2s = ut + \frac{1}{2}at^2 s=(0×5)+12×2×(5)2s = (0 \times 5) + \frac{1}{2} \times 2 \times (5)^2 s=0+1×25=25 ms = 0 + 1 \times 25 = 25 \text{ m}

Explanation:

Since the athlete starts from rest, uu is zero. We plug the known values into the second equation of motion to find the total distance (displacement) traveled.

Problem 2:

A car traveling at 20 m/s20 \text{ m/s} applies brakes and comes to a stop over a distance of 40 m40 \text{ m}. Find the acceleration of the car.

Solution:

Given: u=20 m/su = 20 \text{ m/s}, v=0 m/sv = 0 \text{ m/s}, s=40 ms = 40 \text{ m}. We use the formula: v2=u2+2asv^2 = u^2 + 2as 02=(20)2+2×a×400^2 = (20)^2 + 2 \times a \times 40 0=400+80a0 = 400 + 80a −400=80a-400 = 80a a=−40080=−5 m/s2a = \frac{-400}{80} = -5 \text{ m/s}^2

Explanation:

We use the third equation of motion because time (tt) is not provided. The negative sign in the result indicates that the car is decelerating (slowing down).