Given ( a(t) = \fracdvdt = 6t + 4 ). Integrate: [ v(t) = \int (6t + 4) , dt = 3t^2 + 4t + C_1 ] Using ( v(0)=5 ): ( 5 = 0 + 0 + C_1 \implies C_1 = 5 ). Thus, ( v(t) = 3t^2 + 4t + 5 ).
v(2) = 6(4) – 18(2) + 12 = 24 – 36 + 12 = 0 m/s a(2) = 12(2) – 18 = 24 – 18 = 6 m/s²
| Equation | Description | | :--- | :--- | | ( s = vt ) | Constant velocity motion | | ( v_f = v_i + at ) | Velocity as a function of time | | ( s = v_i t + \frac12 a t^2 ) | Displacement as a function of time | | ( v_f^2 = v_i^2 + 2as ) | Velocity as a function of displacement | rectilinear motion problems and solutions mathalino upd
Miguel smiled. “Mathalino UPD,” he said. “It’s not just answers—it’s a framework. You trace the motion, break it at every change in velocity or acceleration, then rebuild the total journey piece by piece.”
A train accelerates uniformly from rest to a speed of 80 km/h in 10 seconds. Find the acceleration and distance traveled during this time. Given ( a(t) = \fracdvdt = 6t + 4 )
s(t) = 2t³ – 9t² + 12t + 5 v(t) = ds/dt = 6t² – 18t + 12 a(t) = dv/dt = 12t – 18
Compute positions: [ s(0) = 2,\ s(1) = 1 - 6 + 9 + 2 = 6,\ s(3) = 27 - 54 + 27 + 2 = 2,\ s(5) = 125 - 150 + 45 + 2 = 22 ] Displacement = ( s(5) - s(0) = 22 - 2 = 20 ) m (positive, to the right). v(2) = 6(4) – 18(2) + 12 =
Solution:
"It doubles back," Miguel realized. "Then goes forward again. Just like a commuter avoiding traffic."
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.
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