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Particle Motion (Kinematics): Position, Velocity, and Acceleration Explained

Analyzing the movement of a particle along a line using position, velocity, and acceleration.

Core Theorem
Relationship: v(t)=s′(t)v(t) = s'(t) and a(t)=v′(t)a(t) = v'(t). Total distance = ∫ab∣v(t)∣dt\int_{a}^{b} |v(t)| dt. Displacement = ∫abv(t)dt\int_{a}^{b} v(t) dt.
Step-by-Step SOP
  1. 1

    1. Find Roots

    Solve v(t)=0v(t) = 0 to find potential turning points within the given time interval.
  2. 2

    2. Partition the Interval

    Split the original interval [a,b][a, b] into sub-intervals using the roots found in step 1.
  3. 3

    3. Integrate Sub-intervals

    Calculate the definite integral of v(t)v(t) for each sub-interval.
  4. 4

    4. Absolute Sum

    Take the absolute value of each result and add them together to get the total distance.

Practice Exercises

Example 01Easy
A particle moves with velocity v(t)=t2−4v(t) = t^2 - 4 for 0≤t≤30 \le t \le 3. Find the total distance traveled.
NEED A HINT?
Find where v(t)=0v(t) = 0 to check for direction changes. t2−4=0t^2 - 4 = 0 at t=2t=2.
SHOW DETAILED EXPLANATION

Step 1: Find the Turning Point

Set v(t)=0v(t) = 0 to find when the particle stops or changes direction: t2−4=0  ⟹  t=2t^2 - 4 = 0 \implies t = 2. Since t=2t=2 is within the interval [0,3][0, 3], we must split the integral.

Step 2: Determine Direction (Sign of v)

On [0,2][0, 2], v(1)=−3v(1) = -3 (moving left). On [2,3][2, 3], v(2.5)=2.25v(2.5) = 2.25 (moving right). To get distance, we take the absolute value of each movement.

Step 3: Set Up and Solve the Integrals

Total Distance = ∣∫02(t2−4)dt∣+∣∫23(t2−4)dt∣|\int_{0}^{2} (t^2-4)dt| + |\int_{2}^{3} (t^2-4)dt|. The antiderivative is 13t3−4t\frac{1}{3}t^3 - 4t. Evaluated from 0 to 2, we get ∣83−8∣=163|\frac{8}{3}-8| = \frac{16}{3}. From 2 to 3, we get ∣(−3)−(−163)∣=73|(-3) - (-\frac{16}{3})| = \frac{7}{3}.

Step 4: Final Summation

Sum the magnitudes: 163+73=233\frac{16}{3} + \frac{7}{3} = \frac{23}{3}. The total distance is approximately 7.677.67 units.
Common Pitfalls
  • ⚠
    Distance vs. DisplacementDon't just calculate ∫abv(t)dt\int_{a}^{b} v(t) dt. That gives you 'Displacement' (net change in position). Total distance requires the absolute value of velocity.
  • ⚠
    Ignoring the Turning PointsAlways solve v(t)=0v(t)=0 first. If the particle changes direction and you don't split the integral, the 'negative' distance will cancel out the 'positive' distance.
  • ⚠
    Calculation Errors with NegativesWhen evaluating [F(t)]ab[F(t)]_{a}^{b}, be extremely careful with signs like F(b)−(−F(a))F(b) - (-F(a)). Use parentheses to avoid common arithmetic mistakes.
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