Used when acceleration is given as a function of time, position, or velocity ($a = f(t), a = f(s)$, etc.). This requires integration.
Statement:
Velocity of a particle is ( v(t) = t^2 - 4t + 3 ) (m/s). Initial position ( s(0) = 0 ). Find:
Solution:
1. ( s(t) = \int v , dt = \fract^33 - 2t^2 + 3t + C )
( s(0)=0 ) → ( C=0 )
( s(t) = \fract^33 - 2t^2 + 3t )
2. Displacement: ( s(4) = \frac643 - 32 + 12 = \frac643 - 20 = \frac64 - 603 = \frac43 , \textm )
3. Total distance:
Find when ( v(t)=0 ): ( t^2 - 4t + 3 = 0 ) → ( (t-1)(t-3)=0 ) → ( t=1,3 ) rectilinear motion problems and solutions mathalino upd
( s(0) = 0 )
( s(1) = \frac13 - 2 + 3 = \frac13 + 1 = \frac43 )
( s(3) = \frac273 - 18 + 9 = 9 - 9 = 0 )
( s(4) = \frac43 )
Segments:
0→1: ( |4/3 - 0| = 4/3 )
1→3: ( |0 - 4/3| = 4/3 )
3→4: ( |4/3 - 0| = 4/3 )
Total = ( 4/3 + 4/3 + 4/3 = 4 , \textm )
Answers:
( s(t) = \fract^33 - 2t^2 + 3t )
Displacement = ( 4/3 , \textm )
Total distance = ( 4 , \textm )
The search results brought him to a familiar, no-frills webpage. Mathalino wasn’t flashy—no animations, no pop-ups. Just text, equations, and clean line drawings. The site, run by an engineering educator named Romel Verterra, had become a quiet hero for students across the Philippines and beyond.
Miguel found the section: “Rectilinear Motion – Problems and Solutions.” Used when acceleration is given as a function
The page listed problem after problem, from basic to complex. He clicked on one: Problem 1007 – “A car starts from rest and accelerates uniformly…” It showed step-by-step:
He scrolled further. There it was: Problem 1013 – “A particle moves with ( s = t^3 - 6t^2 + 9t ).” Exactly his problem!
The solution was crisp:
Miguel exhaled. It wasn’t just the answer—it was the method. The way Mathalino broke the motion into phases, checking direction changes before integrating absolute values. That was the key he’d missed in lecture.
Problem 3:
The acceleration of a particle moving along a straight line is given bya = 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: 1
Rectilinear motion—the movement of a particle along a straight line—is the cornerstone of engineering mechanics (dynamics). For students at the University of the Philippines Diliman (UPD) and elsewhere, mastering this topic is non-negotiable. Whether you are reviewing for the Engineering Board Exam or tackling your ES 11 (Statics of Rigid Bodies) or ES 12 (Dynamics of Rigid Bodies) homework, you often turn to resources like Mathalino.com for clear, step-by-step solutions.
This article provides a curated collection of rectilinear motion problems and solutions styled after the Mathalino approach. We will cover variable acceleration, constant acceleration, projectile motion (as a special case), and relative motion—all with detailed free-body diagrams (in text form) and algebraic solutions.
📥 Download PDF – 10 additional rectilinear motion problems with solutions
📝 Submit your solution – Join the Mathalino forum for feedback
Let t = time for first stone to hit ground.
Stone 1: y = y₀ + v₀ t + ½ a t²
Take downward positive: y₀=0, y=50 m, v₀=0, a=g=9.81 m/s².
50 = 0 + 0 + ½ (9.81) t² → t² = 100/9.81 → t = √(10.193) ≈ 3.193 s
Stone 2 is thrown 1 second later, so its travel time = t - 1 = 2.193 s.
Initial velocity u (downward positive):
y = u·t₂ + ½ g t₂² → 50 = u(2.193) + ½ (9.81)(2.193)²
½(9.81)(4.809) = 23.58
Thus 50 = 2.193u + 23.58 → 2.193u = 26.42 → u ≈ 12.04 m/s downward.
✅ Answer: The second stone’s initial velocity is 12.04 m/s downward.