Rectilinear Motion Problems And Solutions Mathalino Upd |link|

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    Rectilinear Motion Problems And Solutions Mathalino Upd |link|

    Rectilinear Motion Problems and Solutions: A Comprehensive Guide (Mathalino UPD Style)

    Solution (Mathalino approach – use relative motion or equal position):

    Let s=0 at Car B’s initial position.
    For Car A: s_A = 100 + 20t (since 100 m ahead at t=0, vel=20)
    For Car B: s_B = 0 + 0·t + ½ (2) t² = t²

    Overtaking when s_B = s_A:
    t² = 100 + 20t → t² - 20t - 100 = 0
    Solve: t = [20 ± √(400 + 400)]/2 = [20 ± √800]/2 = [20 ± 28.284]/2
    Positive root: t = (48.284)/2 = 24.142 s

    ✅ Answer: Car B overtakes after 24.14 seconds.

    Solution (Step-by-Step):

    Step 1: Differentiate to get v(t) and a(t).
    s(t) = 2t³ – 9t² + 12t + 5
    v(t) = ds/dt = 6t² – 18t + 12
    a(t) = dv/dt = 12t – 18 rectilinear motion problems and solutions mathalino upd

    Step 2: Evaluate at t = 2 s.
    v(2) = 6(4) – 18(2) + 12 = 24 – 36 + 12 = 0 m/s
    a(2) = 12(2) – 18 = 24 – 18 = 6 m/s²

    Step 3: Particle at rest means v(t) = 0.
    6t² – 18t + 12 = 0 → divide 6: t² – 3t + 2 = 0 → (t-1)(t-2) = 0
    Thus at t = 1 s and t = 2 s, the particle is momentarily at rest.

    Step 4: Total distance traveled (t=0 to t=4).
    We check the sign of velocity in intervals [0,1], [1,2], [2,4]. Solution (Step-by-Step): Step 1: Differentiate to get v(t)

    Compute positions: s(0)=5 m
    s(1)=2-9+12+5=10 m
    s(2)=16-36+24+5=9 m
    s(4)=128-144+48+5=37 m

    Distance:
    From 0→1: |10-5| = 5 m
    From 1→2: |9-10| = 1 m
    From 2→4: |37-9| = 28 m
    Total distance = 5 + 1 + 28 = 34 m.

    ✅ Answer: (a) v=0, a=6 m/s²; (b) t=1 s, 2 s; (c) 34 m. At t=0: v=12 m/s (+) At t=1

    Scenario 3: Variable Acceleration (Integration)

    Problem: The acceleration is not constant but depends on time or velocity.

    Example: A particle moves along a straight line such that its acceleration is $a = (2t - 4) , \textm/s^2$. If $v = 0$ and $s = 0$ when $t = 0$, find the velocity and position at $t = 3$ seconds.

    Solution:

    1. Find Velocity: Integrate acceleration. $$v = \int a , dt = \int (2t - 4) , dt = t^2 - 4t + C_1$$ At $t=0, v=0 \implies C_1 = 0$. $$v = t^2 - 4t$$ At $t=3$: $v = 3^2 - 4(3) = 9 - 12 = -3 , \textm/s$.

    2. Find Position: Integrate velocity. $$s = \int v , dt = \int (t^2 - 4t) , dt = \fract^33 - 2t^2 + C_2$$ At $t=0, s=0 \implies C_2 = 0$. $$s = \fract^33 - 2t^2$$ At $t=3$: $s = \frac273 - 2(9) = 9 - 18 = -9 , \textm$.


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