Rectilinear Motion Problems And Solutions Mathalino Upd -

, refers to the movement of a particle along a straight line

. It is categorized into three main types based on acceleration Uniform Motion: Constant velocity ( Uniformly Accelerated Motion: Constant acceleration ( Variable Acceleration: Acceleration changes over time ( Core Formulas for Rectilinear Translation

For any rectilinear problem involving constant acceleration, these fundamental equations apply Velocity-Time: Displacement-Time: Velocity-Displacement: Free-Falling Bodies , simply replace acceleration ( ) with gravity ( for downward motion and for upward motion Sample Problems and Solutions 1. The "Return in 10 Seconds" Problem

A stone is thrown vertically upward and returns to earth in 10 seconds. Find its initial velocity and maximum height The total time is 10 seconds, meaning it takes to reach the peak and 5 seconds to fall back . At the peak, final velocity ( ) is zero. Initial Velocity (

v sub f equals v sub i minus g t ⟹ 0 equals v sub i minus 9.81 open paren 5 close paren ⟹ v sub i equals 49.05 space m/s Maximum Height (

h equals one-half g t squared ⟹ h equals one-half open paren 9.81 close paren open paren 5 squared close paren ⟹ h equals 122.625 space m 2. Meeting Stones in Mid-Air

A stone is dropped from a 1000 ft balloon. Two seconds later, another stone is thrown upward from the ground at 248 ft/s. When and where do they pass each other be the time for the first stone. The second stone's time is Stone 1 (Falling): Stone 2 (Rising): (total height): rectilinear motion problems and solutions mathalino upd

16 t sub 1 squared plus open bracket 248 open paren t sub 1 minus 2 close paren minus 16 open paren t sub 1 minus 2 close paren squared close bracket equals 1000 Solving this yields They pass at (or approx. 600 ft) above the ground 3. Constant Deceleration (The Train Problem)

A train travels 24 ft during its 10th second and 18 ft during its 12th second. Find its initial velocity and acceleration

Treat the distance in a specific second as the instantaneous velocity at the midpoint of that second ( Subtracting (2) from (1): Plugging back: For more complex challenges involving Variable Acceleration Moving Vessels , visit the full MATHalino Kinematics Review problem involving calculus? Kinematics | Engineering Mechanics Review at MATHalino

Note: • is positive (+) if is increasing (accelerate). (decelerate). • is positive (+) if the particle is moving downward. Kinematics | Engineering Mechanics Review at MATHalino

Rectilinear motion (or rectilinear translation) refers to the movement of a particle along a single straight-line path. At MATHalino, this topic is a core component of Engineering Mechanics (Dynamics), covering everything from uniform velocity to variable acceleration. Core Formulas for Rectilinear Motion

Problem solving at MATHalino generally falls into three categories based on acceleration: Motion Type Governing Equations Context/Usage Uniform Motion Constant velocity; zero acceleration. Constant Acceleration Used for cars braking or free-falling bodies ( Variable Acceleration Requires calculus (differentiation or integration). Featured Problems & Solutions (MATHalino) , refers to the movement of a particle along a straight line

These examples represent common problem types found in the MATHalino Rectilinear Translation Reviewer: Kinematics | Engineering Mechanics Review at MATHalino

Introduction

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.

Rectilinear Motion Problems and Solutions: A Mathalino-Inspired Guide for Engineering Students (UPD)

Problem 1: Constant Acceleration (Overtaking)

Statement:
A car starts from rest and accelerates at ( 2 , \textm/s^2 ). At the same instant, a truck moving at constant speed ( 10 , \textm/s ) overtakes the car. How long will it take for the car to catch up with the truck, and how far will the car have traveled?

Solution:

Let ( t = 0 ) be the start.
Car: ( s_c = 0 + 0 \cdot t + \frac12 (2) t^2 = t^2 )
Truck: ( s_t = 10t ) Solution: Let t = time for first stone to hit ground

Catch up: ( s_c = s_t )
( t^2 = 10t )
( t(t - 10) = 0 ) → ( t = 10 , \texts ) (ignore ( t=0 ))

Distance: ( s = t^2 = 100 , \textm )

Answer: ( t = 10 , \texts, \quad s = 100 , \textm )

Solution:

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.