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Motion, Forces and Energy - Motion

Grade 12A LevelPhysics

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

🔑Concepts

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Distance is a scalar quantity representing the total path length, while Displacement is a vector quantity representing the straight-line change in position from a start point to an end point, denoted as Δs\Delta s.

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Average Speed is calculated as the total distance divided by total time (v=dtv = \frac{d}{t}), whereas Velocity is the rate of change of displacement (v=ΔsΔtv = \frac{\Delta s}{\Delta t}).

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Acceleration (aa) is a vector quantity defined as the rate of change of velocity. If an object slows down, it experiences negative acceleration, also known as deceleration.

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On a Displacement-Time graph, the gradient (slope) at any point represents the velocity. A curved line indicates changing velocity (acceleration).

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On a Velocity-Time graph, the gradient represents the acceleration (aa), and the area under the graph represents the total displacement (ss).

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Uniformly accelerated motion can be described using the kinematic equations (SUVAT), provided that acceleration remains constant.

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Objects in Free Fall near the Earth's surface accelerate at a constant rate g≈9.81 m/s2g \approx 9.81\,m/s^2 (often taken as 10 m/s210\,m/s^2 in IGCSE) in the absence of air resistance.

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Terminal Velocity is reached when the upward force of air resistance equals the downward force of weight (W=mgW = mg), resulting in a net force of 0 N0\,N and zero acceleration.

📐Formulae

v=stv = \frac{s}{t}

a=v−uta = \frac{v - u}{t}

v=u+atv = u + at

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

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

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

💡Examples

Problem 1:

A car travelling at 15 m/s15\,m/s accelerates uniformly at a rate of 2 m/s22\,m/s^2 for a period of 6 s6\,s. Calculate the final velocity of the car and the distance it travels during this acceleration.

Solution:

Given: u=15 m/su = 15\,m/s, a=2 m/s2a = 2\,m/s^2, t=6 st = 6\,s.

  1. To find vv: v=u+at=15+(2×6)=27 m/sv = u + at = 15 + (2 \times 6) = 27\,m/s.
  2. To find ss: s=ut+12at2=(15×6)+12(2)(62)=90+36=126 ms = ut + \frac{1}{2}at^2 = (15 \times 6) + \frac{1}{2}(2)(6^2) = 90 + 36 = 126\,m.

Explanation:

We use the equations of motion for constant acceleration. First, we find the final velocity using the first SUVAT equation, then use the displacement equation to find the total distance covered.

Problem 2:

A stone is dropped from the top of a cliff and hits the ground after 4.0 s4.0\,s. Taking g=9.8 m/s2g = 9.8\,m/s^2 and ignoring air resistance, calculate the height of the cliff.

Solution:

Given: u=0 m/su = 0\,m/s (dropped from rest), t=4.0 st = 4.0\,s, a=g=9.8 m/s2a = g = 9.8\,m/s^2. Using s=ut+12at2s = ut + \frac{1}{2}at^2: s=(0×4.0)+12(9.8)(4.02)s = (0 \times 4.0) + \frac{1}{2}(9.8)(4.0^2) s=0+4.9×16=78.4 ms = 0 + 4.9 \times 16 = 78.4\,m.

Explanation:

Since the stone is 'dropped', the initial velocity is zero. We apply the displacement formula with gravity as the acceleration to find the vertical distance traveled, which is the height of the cliff.