And Solutions Mathalino Upd — Rectilinear Motion Problems

Below are representative problems frequently found in MATHalino’s Engineering Mechanics archives:

Vertical Projection (Problem 1003): A stone thrown vertically upward returns in 10 seconds.

Solution Strategy: Total time is split equally (5s up, 5s down). Using for the upward trip ( ), initial velocity is calculated as . Max height (

Relative Velocity (Problem 1004): A ball is dropped from an 80 ft tower as another is thrown up from the ground at 40 ft/s.

Solution Strategy: Set the sum of their displacements equal to the tower height ( ). Solving for shows they pass after 2 seconds.

Variable Acceleration: Problems where acceleration is a function of time (

) require integration of the acceleration function to find velocity and position. 4. Problem Solving Procedure To solve these problems systematically, follow these steps:

Rectilinear Motion of Particles: Formulas, Examples & Key Concepts

Statement:
A particle moves along a straight line such that its position is given by
[ s(t) = t^3 - 6t^2 + 9t + 2 ]
where ( s ) is in meters, ( t ) in seconds. Find:

Solution:

1. Velocity & acceleration
( v(t) = 3t^2 - 12t + 9 )
( a(t) = 6t - 12 )

At ( t = 2 ): ( v = 3(4) - 24 + 9 = -3 , \textm/s )
( a = 6(2) - 12 = 0 , \textm/s^2 )

2. At rest: ( v=0 ) → ( 3t^2 - 12t + 9 = 0 ) → divide 3: ( t^2 - 4t + 3 = 0 ) → ( (t-1)(t-3)=0 )
( t = 1 , \texts ) and ( t = 3 , \texts )

3. Displacement (0 to 4):
( s(0) = 2 )
( s(4) = 64 - 96 + 36 + 2 = 6 )
Displacement = ( s(4) - s(0) = 6 - 2 = 4 , \textm )

4. Total distance:
Check direction changes at ( t=1,3 ).
( s(1) = 1 - 6 + 9 + 2 = 6 )
( s(3) = 27 - 54 + 27 + 2 = 2 )

Segments:
0→1: ( |6-2| = 4 , \textm )
1→3: ( |2-6| = 4 , \textm )
3→4: ( |6-2| = 4 , \textm )
Total = ( 4+4+4 = 12 , \textm )

Answers:
( v(2) = -3 , \textm/s, a(2) = 0 )
At rest at ( t = 1, 3 ) s
Displacement = ( 4 , \textm )
Distance = ( 12 , \textm )

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. rectilinear motion problems and solutions mathalino upd

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. Statement: A particle moves along a straight line

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.