Upd — Rectilinear Motion Problems And Solutions Mathalino

It was 11:47 PM. The air in the cramped dorm room smelled of instant coffee and desperate ambition. Miguel, a second-year civil engineering student at the University of the Philippines Diliman (UPD), stared at his problem set. On the page, a single sentence mocked him:

“A particle moves along a straight line according to the equation ( s = t^3 - 6t^2 + 9t ), where ( s ) is in meters and ( t ) in seconds. Find the total distance traveled from ( t=0 ) to ( t=5 ) seconds.”

Miguel’s hand trembled. He knew the theory: displacement, velocity, acceleration, time intervals. But applying it? That required a systematic method—one his professor assumed they already mastered. His classmates had mentioned a website: Mathalino. “Just search ‘rectilinear motion problems and solutions,’” they said. “It’s a goldmine.”

He opened his laptop. The screen’s glow illuminated his tired eyes. He typed: rectilinear motion problems and solutions mathalino.

Problem 2 (UPD falling egg problem):
A student drops a stone from a 50-meter-high dormitory roof at UPD. One second later, he throws another stone vertically downward from the same point. Both stones hit the ground at the same time. What was the initial velocity of the second stone? (Use g = 9.81 m/s²)

Over the next week, Miguel used the updated Mathalino to solve every rectilinear problem in his syllabus. He learned to handle:

By the time the long exam arrived, Miguel no longer feared phrases like “rectilinear motion with variable acceleration” or “distance vs displacement.” He even corrected the professor’s typo on a sample problem (the prof had forgotten a sign change at a turning point).

After the exam, his classmates gathered around. “How’d you get the last problem? The one with the ball rolling down a track then onto a flat surface?”

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.”

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.

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. rectilinear motion problems and solutions mathalino upd

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.

Statement:
A particle moves with position ( s(t) = 2\sin(3t) ), ( t ) in seconds, ( s ) in meters. Find:

Solution:

1. ( v(t) = 6\cos(3t) )
( a(t) = -18\sin(3t) )

2. Max speed = max ( |v(t)| = |6\cos(3t)| = 6 , \textm/s ) (since max ( |\cos| = 1 )).

3. ( a(t) = 0 ) → ( -18\sin(3t) = 0 ) → ( \sin(3t) = 0 ) → ( 3t = n\pi ) → ( t = \fracn\pi3 )
Smallest positive ( t ): ( n=1 ) → ( t = \pi/3 \approx 1.047 , \texts )

Answers:
( v(t) = 6\cos(3t) ), ( a(t) = -18\sin(3t) )
Max speed = ( 6 , \textm/s )
First zero acceleration at ( t = \pi/3 , \texts )

Diliman, Quezon City – 11:47 PM

Miguel stared at his laptop screen. The tab open said Mathalino | Rectilinear Motion Problems. Another tab read UPD CVS (College of Science complex map). He was a third-year mechanical engineering student, and in six hours, he had his final exam in ES 11 (Dynamics of Rigid Bodies).

He wasn’t worried about the theories. He was worried about the twist.

His professor, Dr. Reyes, was famous for one thing: giving a rectilinear motion problem that looked like a simple “a = dv/dt” exercise, but was actually a philosophical trap.

Miguel scrolled through Mathalino’s solved problems. Problem 01: A car accelerates from rest… Too easy. Problem 15: A particle moves along a straight line with v = t^2 – 4t + 3… He could do that in his sleep. It was 11:47 PM

Then he saw it. A problem titled: “Updated: UPD Engineering – Rectilinear Motion with Two-Way Constraints.”

The problem read:

A jeepney traveling along University Avenue from the Philcoa gate suddenly breaks down 200 meters before the Vinzons Hall stop. A student, late for class, runs from the jeepney toward Vinzons at a constant velocity of 3 m/s. At the same instant, a second student on a bike leaves Vinzons Hall heading toward the jeepney with an initial velocity of 2 m/s and accelerates at 0.5 m/s². When and where do they meet? Assume rectilinear motion along a straight path.

Miguel grabbed his yellow pad. He set the origin at the jeepney’s breakdown point, positive direction toward Vinzons. Distance between them: 200 m.

For the runner (constant velocity):
( x_1 = 3t )

For the biker (accelerating):
( x_2 = 200 - (2t + 0.5 \cdot 0.5 t^2) )
Wait—he paused. Careful. The biker starts at ( x = 200 ) and moves toward decreasing x.

Meeting condition: ( x_1 = x_2 )
( 3t = 200 - 2t - 0.25 t^2 )
( 0.25 t^2 + 5t - 200 = 0 )
Multiply by 4: ( t^2 + 20t - 800 = 0 )
( t = \frac-20 \pm \sqrt400 + 32002 = \frac-20 \pm 602 )
Positive root: ( t = 20 ) seconds.

Then ( x = 3(20) = 60 ) meters from the jeepney.

He checked Mathalino’s solution. Correct. Simple quadratic. But that wasn’t the twist.

He refreshed the page. The problem updated again—probably a glitch or an Easter egg left by some former Isko.

The new text read:

Bonus: Suppose the runner’s velocity is not constant but v_r = 3 – 0.1t (m/s) due to fatigue, and the biker’s acceleration stops at t = 10 s, after which velocity is constant. Solve for meeting time. “A particle moves along a straight line according

Miguel grinned. That was the infamous UPD twist—real-world fatigue and mechanical limits.

He worked through it in two phases. Phase 1 (0 to 10 s):

At t=10:
( x_r1 = 30 – 5 = 25 ) m
( x_b1 = 200 – (20 + 25) = 155 ) m → separation = 130 m.

Phase 2 (t > 10 s): Runner: ( v_r = 3 – 1 = 2 ) m/s constant.
Biker: ( v_b = 2 + 5 = 7 ) m/s constant.

Relative speed = 2 + 7 = 9 m/s closing.
Time to close 130 m = 130/9 ≈ 14.44 s.

Total time = 10 + 14.44 = 24.44 s.
Meeting point from jeepney: ( x = 25 + 2(14.44) = 53.89 ) m.

He compared with Mathalino’s hidden solution. Match.

Miguel leaned back. Rectilinear motion wasn’t just about formulas—it was about when to switch equations, when reality breaks the ideal case. That’s why UPD engineers fear and love it.

At 4:00 AM, he closed the laptop. He didn’t memorize solutions. He understood the motion.

That morning, Dr. Reyes gave a problem about a train, a student, and a variable acceleration. Miguel finished in 20 minutes.

On his paper, he wrote: “Solution courtesy of Mathalino and a sleepless night at UPD.”

He passed.

End of story.
Need a problem set based on this story or a derivation of the equations used? Just ask.