Rectilinear Motion Problems And Solutions Mathalino Upd Patched

✅ Answer: The second stone’s initial velocity is . 4. Problem Set #3: Motion with Variable Acceleration (Integration) Problem 3: The acceleration of a particle moving along a straight line is given by a = 4 - t² (in m/s²). At t=0, v=3 m/s and s=2 m. Find (a) v as a function of t, (b) s as a function of t, (c) the velocity when t=4 s, and (d) the displacement from t=0 to t=4 s. Solution: a) Velocity: a = dv/dt = 4 - t² → dv = (4 - t²) dt Integrate: v(t) = ∫(4 - t²) dt = 4t - t³/3 + C At t=0, v=3 → 3 = 0 - 0 + C → C=3. Thus v(t) = 4t - t³/3 + 3 m/s.

s(4) = 2(16) - (256)/12 + 3(4) + 2 = 32 - 21.333 + 12 + 2 = 24.667 m s(0)=2 m → Displacement = 24.667 - 2 = 22.667 m . rectilinear motion problems and solutions mathalino upd

Distance: From 0→1: |10-5| = 5 m From 1→2: |9-10| = 1 m From 2→4: |37-9| = 28 m ✅ Answer: The second stone’s initial velocity is