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Space, Time and Motion - Kinematics

Grade 12IBPhysics

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

🔑Concepts

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Displacement (ss) is a vector quantity representing the change in position, while distance is a scalar representing the total path length.

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Velocity (vv) is the rate of change of displacement (v=dsdtv = \frac{ds}{dt}), and Acceleration (aa) is the rate of change of velocity (a=dvdta = \frac{dv}{dt}).

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

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In a velocity-time (v−tv-t) graph, the gradient represents the acceleration, and the area under the curve represents the displacement (Δs\Delta s).

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Equations of motion (SUVAT) are only applicable when acceleration (aa) is constant.

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Projectile motion involves independent horizontal and vertical components. In the absence of air resistance, horizontal velocity (vxv_x) remains constant while vertical motion is subject to constant gravitational acceleration (g≈9.81 m s−2g \approx 9.81 \text{ m s}^{-2}).

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Fluid resistance (drag) acts opposite to the direction of motion and typically increases with speed, eventually leading to terminal velocity where Fdrag=FweightF_{drag} = F_{weight} and a=0a = 0.

📐Formulae

v=u+atv = u + at

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

s=(u+v)t2s = \frac{(u + v)t}{2}

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

s=vt−12at2s = vt - \frac{1}{2}at^2

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

a=ΔvΔta = \frac{\Delta v}{\Delta t}

💡Examples

Problem 1:

A ball is kicked from the ground with an initial velocity of 20 m s−120 \text{ m s}^{-1} at an angle of 30∘30^\circ to the horizontal. Calculate the maximum height reached by the ball. (Take g=9.81 m s−2g = 9.81 \text{ m s}^{-2})

Solution:

  1. Identify the vertical component of the initial velocity: uy=usin⁡(θ)=20sin⁡(30∘)=10 m s−1u_y = u \sin(\theta) = 20 \sin(30^\circ) = 10 \text{ m s}^{-1}.
  2. At maximum height, the vertical velocity vy=0 m s−1v_y = 0 \text{ m s}^{-1}.
  3. Use the formula vy2=uy2+2aysv_y^2 = u_y^2 + 2a_y s: 02=(10)2+2(−9.81)s0^2 = (10)^2 + 2(-9.81)s 0=100−19.62s0 = 100 - 19.62s s=10019.62≈5.10 ms = \frac{100}{19.62} \approx 5.10 \text{ m}

Explanation:

To find the maximum height, we focus solely on the vertical component of motion. We use the kinematic equation relating velocity, acceleration, and displacement, setting the final vertical velocity to zero at the peak.

Problem 2:

A car accelerates uniformly from rest to a speed of 30 m s−130 \text{ m s}^{-1} over a distance of 150 m150 \text{ m}. Determine the acceleration of the car.

Solution:

Given: u=0u = 0, v=30 m s−1v = 30 \text{ m s}^{-1}, s=150 ms = 150 \text{ m}. We need to find aa. Using v2=u2+2asv^2 = u^2 + 2as: (30)2=(0)2+2(a)(150)(30)^2 = (0)^2 + 2(a)(150) 900=300a900 = 300a a=900300=3.0 m s−2a = \frac{900}{300} = 3.0 \text{ m s}^{-2}

Explanation:

Since the acceleration is uniform, we can apply the SUVAT equations. We select the equation that excludes time (tt) because it was not provided in the problem statement.

Kinematics - Revision Notes & Key Formulas | IB Grade 12 Physics