Rectilinear motion—the movement of a particle along a straight line—is one of the most fundamental topics in differential and integral calculus. For engineering students, particularly those from the University of the Philippines Diliman (UPD) and readers of the renowned Mathalino online community, mastering this topic is non-negotiable. It forms the backbone of dynamics, physics, and even structural engineering.
In this article, we will dissect rectilinear motion problems and solutions using the classic Mathalino approach: rigorous derivation, step-by-step solutions, and real-world engineering problems. We will cover the core relationships between position, velocity, acceleration, and time, followed by solved problems that mirror the difficulty of UPD’s Engineering Math exams.
Last updated: April 12, 2026
Tag: Dynamics, Engineering Mechanics, Calculus
Rectilinear motion problems and solutions are the training ground for logical thinking in engineering. Resources like Mathalino provide excellent problem collections, and this guide—tailored for UPD students—offers a dynamic, integrated approach to mastering them. Practice regularly, draw diagrams, and always check the physical plausibility of your answer.
For more problems, visit the Mathalino website’s Rectilinear Motion section or consult Engineering Mechanics: Dynamics by Hibbeler. If you’re an UPD student, work with your ES 12 instructors and use past quizzes for practice.
Need more solutions? Leave a comment or request specific rectilinear motion problems below!
Keywords: rectilinear motion problems and solutions mathalino upd, dynamics problem set, particle kinematics, engineering mechanics review.
| Problem | Key Result | | --- | --- | | 1 | ( t = 10 , \texts, s = 100 , \textm ) | | 2 | Total distance = 12 m | | 3 | No finite max velocity | | 4 | Max speed = 6 m/s | | 5 | Distance = 4 m | rectilinear motion problems and solutions mathalino upd
Statement: The velocity of a particle is ( v(t) = 2t - 4 ) m/s for ( 0 \le t \le 6 ). Find the total distance traveled.
Solution (without graphing):
Find when ( v(t)=0 ): ( 2t-4=0 \implies t=2 ) s.
Displacement from t=0 to t=2: [ \int_0^2 (2t-4) dt = [t^2 - 4t]_0^2 = (4-8) - 0 = -4 \ \textm ] Distance part 1 = ( | -4 | = 4 ) m.
Displacement from t=2 to t=6: [ \int_2^6 (2t-4) dt = [t^2 - 4t]_2^6 = (36-24) - (4-8) = 12 - (-4) = 16 \ \textm ] Distance part 2 = ( 16 ) m (positive, no absolute needed).
Total distance = ( 4 + 16 = 20 ) m.
Answer: 20 meters.
Statement:
The acceleration of a particle is given by ( a(t) = 12t - 6 ). At ( t=0 ), ( v_0 = 5 , \textm/s ), ( s_0 = 2 , \textm ). Find:
Solution:
1. Velocity:
( v(t) = \int a , dt = \int (12t - 6) dt = 6t^2 - 6t + C )
Using ( v(0) = 5 ): ( C = 5 )
( v(t) = 6t^2 - 6t + 5 )
2. Position:
( s(t) = \int v , dt = \int (6t^2 - 6t + 5) dt = 2t^3 - 3t^2 + 5t + D )
Using ( s(0) = 2 ): ( D = 2 )
( s(t) = 2t^3 - 3t^2 + 5t + 2 )
3. Max velocity:
Set ( a(t) = 0 ) → ( 12t - 6 = 0 ) → ( t = 0.5 , \texts )
Check second derivative of ( v ): ( v'(t) = a(t) ), ( a'(t) = 12 > 0 ) → minimum actually (since concave up)
Wait — ( a(t) = 12t - 6 ), derivative of ( a ) = 12 > 0 → acceleration increasing, so ( v ) has minimum at ( t=0.5 ).
Thus, no maximum for ( t \ge 0 ) — velocity increases indefinitely. So answer: no max (or infinite).
Answers:
( v(t) = 6t^2 - 6t + 5 )
( s(t) = 2t^3 - 3t^2 + 5t + 2 )
No finite maximum velocity.
Statement:
Velocity of a particle is ( v(t) = t^2 - 4t + 3 ) (m/s). Initial position ( s(0) = 0 ). Find: Rectilinear motion—the movement of a particle along a
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 )
( 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 ) Need more solutions
In Kinematics, there are three distinct ways to solve rectilinear motion problems depending on the variables given. Identifying the missing variable is the key to selecting the correct equation.
Let:
