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Calculus - Kinematics

Grade 11IB_AA

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

🔑Concepts

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Kinematics involves the study of the motion of particles. In Calculus, we relate position, velocity, and acceleration using derivatives and integrals.

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Position s(t)s(t) represents the location of a particle at time tt relative to a fixed origin. Displacement is the change in position over a specific time interval, given by s(t2)−s(t1)s(t_2) - s(t_1).

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Velocity v(t)v(t) is the rate of change of position with respect to time: v(t)=s′(t)v(t) = s'(t). If v(t)>0v(t) > 0, the particle is moving in the positive direction; if v(t)<0v(t) < 0, it moves in the negative direction.

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Acceleration a(t)a(t) is the rate of change of velocity with respect to time: a(t)=v′(t)=s′′(t)a(t) = v'(t) = s''(t). A particle is speeding up if v(t)v(t) and a(t)a(t) have the same sign, and slowing down if they have opposite signs.

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A particle is at 'instantaneous rest' when v(t)=0v(t) = 0. The direction of motion changes when v(t)v(t) crosses the tt-axis (changes sign).

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Speed is the magnitude of velocity: Speed=∣v(t)∣\text{Speed} = |v(t)|. Total distance traveled is the integral of speed over an interval: ∫t1t2∣v(t)∣dt\int_{t_1}^{t_2} |v(t)| dt.

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To find velocity from acceleration or position from velocity, we use integration: v(t)=∫a(t)dtv(t) = \int a(t) dt and s(t)=∫v(t)dts(t) = \int v(t) dt. Don't forget the constant of integration CC, which is usually determined by 'initial conditions' at t=0t = 0.

📐Formulae

v(t)=dsdt=s′(t)v(t) = \frac{ds}{dt} = s'(t) caviar

a(t)=dvdt=v′(t)=s′′(t)a(t) = \frac{dv}{dt} = v'(t) = s''(t) caviar

Displacement from t1 to t2=∫t1t2v(t)dt=s(t2)−s(t1)\text{Displacement from } t_1 \text{ to } t_2 = \int_{t_1}^{t_2} v(t) dt = s(t_2) - s(t_1) caviar

Total Distance from t1 to t2=∫t1t2∣v(t)∣dt\text{Total Distance from } t_1 \text{ to } t_2 = \int_{t_1}^{t_2} |v(t)| dt caviar

s(t)=∫v(t)dt+Cs(t) = \int v(t) dt + C caviar

v(t)=∫a(t)dt+Cv(t) = \int a(t) dt + C caviar

💡Examples

Problem 1:

A particle moves in a straight line such that its position ss (in meters) at time tt (in seconds) is given by s(t)=t3−9t2+24ts(t) = t^3 - 9t^2 + 24t for t≥0t \ge 0. Find the time(s) when the particle is at rest and find the acceleration at those times.

Solution:

  1. Find the velocity function by differentiating position: v(t)=s′(t)=3t2−18t+24v(t) = s'(t) = 3t^2 - 18t + 24
  2. Set velocity to zero to find when the particle is at rest: 3t2−18t+24=03t^2 - 18t + 24 = 0 3(t2−6t+8)=03(t^2 - 6t + 8) = 0 3(t−2)(t−4)=03(t - 2)(t - 4) = 0 The particle is at rest at t=2t = 2 and t=4t = 4 seconds.
  3. Find the acceleration function by differentiating velocity: a(t)=v′(t)=6t−18a(t) = v'(t) = 6t - 18
  4. Calculate acceleration at the specific times: At t=2t = 2: a(2)=6(2)−18=−6 m/s2a(2) = 6(2) - 18 = -6 \text{ m/s}^2 At t=4t = 4: a(4)=6(4)−18=6 m/s2a(4) = 6(4) - 18 = 6 \text{ m/s}^2

Explanation:

To find rest points, we solve v(t)=0v(t)=0. The acceleration is the second derivative of the position function.

Problem 2:

A particle's velocity is given by v(t)=6t−12v(t) = 6t - 12 m/s. Find the total distance traveled by the particle in the first 3 seconds.

Solution:

  1. Determine if the particle changes direction in the interval [0,3][0, 3]: v(t)=0  ⟹  6t−12=0  ⟹  t=2v(t) = 0 \implies 6t - 12 = 0 \implies t = 2
  2. The velocity is negative for 0≤t<20 \le t < 2 and positive for 2<t≤32 < t \le 3.
  3. Calculate the total distance as the sum of absolute displacements: Total Distance=∫03∣6t−12∣dt\text{Total Distance} = \int_{0}^{3} |6t - 12| dt Total Distance=∣∫02(6t−12)dt∣+∣∫23(6t−12)dt∣\text{Total Distance} = \left| \int_{0}^{2} (6t - 12) dt \right| + \left| \int_{2}^{3} (6t - 12) dt \right|
  4. Evaluate the integrals: ∫(6t−12)dt=3t2−12t\int (6t - 12) dt = 3t^2 - 12t Interval 1: [3t2−12t]02=(3(4)−12(2))−0=12−24=−12[3t^2 - 12t]_0^2 = (3(4) - 12(2)) - 0 = 12 - 24 = -12 Interval 2: [3t2−12t]23=(3(9)−12(3))−(−12)=(27−36)+12=−9+12=3[3t^2 - 12t]_2^3 = (3(9) - 12(3)) - (-12) = (27 - 36) + 12 = -9 + 12 = 3
  5. Total distance = ∣−12∣+∣3∣=12+3=15|-12| + |3| = 12 + 3 = 15 meters.

Explanation:

Because velocity changes sign at t=2t=2, we must split the integral into two parts to find the total distance, otherwise the negative and positive displacements would partially cancel out.