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Space, Time and Motion - Kinematics

Grade 11IBPhysics

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 (s⃗\vec{s}) is a vector representing the change in position.

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Speed is the rate of change of distance (scalar), whereas velocity (v⃗\vec{v}) is the rate of change of displacement (vector).

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Acceleration (a⃗\vec{a}) is defined as the rate of change of velocity: a=ΔvΔta = \frac{\Delta v}{\Delta t}. It is measured in m s−2\text{m s}^{-2}.

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For motion with constant acceleration, the SUVAT equations are used, where ss = displacement, uu = initial velocity, vv = final velocity, aa = acceleration, and tt = time.

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On a displacement-time (s−ts-t) graph, the gradient represents the instantaneous velocity.

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On a velocity-time (v−tv-t) graph, the gradient represents acceleration, and the area under the curve represents the displacement.

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Projectile motion treats horizontal and vertical components independently. Horizontal acceleration is 0 m s−20 \text{ m s}^{-2} (ignoring air resistance), and vertical acceleration is g=−9.81 m s−2g = -9.81 \text{ m s}^{-2}.

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Fluid resistance (drag) acts opposite to the direction of motion and increases with speed, eventually leading to terminal velocity where a=0a = 0.

📐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)t2s = \frac{(u+v)t}{2}

s=vt−12at2s = vt - \frac{1}{2}at^2

vavg=ΔsΔtv_{avg} = \frac{\Delta s}{\Delta t}

💡Examples

Problem 1:

A car accelerates uniformly from rest to a velocity of 20 m s−120 \text{ m s}^{-1} over a distance of 50 m50 \text{ m}. Calculate the acceleration of the car.

Solution:

Using the formula v2=u2+2asv^2 = u^2 + 2as: (20)2=(0)2+2(a)(50)(20)^2 = (0)^2 + 2(a)(50) 400=100a400 = 100a a=4 m s−2a = 4 \text{ m s}^{-2}

Explanation:

We identify the knowns: u=0u=0, v=20v=20, s=50s=50. Since time tt is not provided, we use the SUVAT equation that relates v,u,a,v, u, a, and ss.

Problem 2:

A ball is thrown horizontally from a cliff of height 45 m45 \text{ m} with an initial speed of 15 m s−115 \text{ m s}^{-1}. How far from the base of the cliff does the ball land?

Solution:

First, find time tt using vertical motion (uy=0u_y = 0, sy=−45 ms_y = -45 \text{ m}, g=−9.81 m s−2g = -9.81 \text{ m s}^{-2}): sy=uyt+12at2  ⟹  −45=0+12(−9.81)t2s_y = u_y t + \frac{1}{2}at^2 \implies -45 = 0 + \frac{1}{2}(-9.81)t^2 t=2×459.81≈3.03 st = \sqrt{\frac{2 \times 45}{9.81}} \approx 3.03 \text{ s} Then, find horizontal range sxs_x: sx=ux×t=15×3.03=45.45 ms_x = u_x \times t = 15 \times 3.03 = 45.45 \text{ m}

Explanation:

In projectile motion, vertical displacement determines the time of flight, which is then multiplied by the constant horizontal velocity to find the range.

Kinematics - Revision Notes & Key Formulas | IB Grade 11 Physics