Rectilinear Motion Problems And Solutions | Mathalino Upd

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.”

Problem: The acceleration is not constant but depends on time or velocity.

Example: A particle moves along a straight line such that its acceleration is $a = (2t - 4) , \textm/s^2$. If $v = 0$ and $s = 0$ when $t = 0$, find the velocity and position at $t = 3$ seconds. rectilinear motion problems and solutions mathalino upd

Solution:

When solving problems similar to those found in Mathalino or Board Exams:

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. Over the next week, Miguel used the updated

Statement: A particle moves along a straight line such that its position is defined by ( s(t) = t^3 - 6t^2 + 9t + 2 ) meters, where ( t ) is in seconds. Determine: (a) Velocity and acceleration at ( t = 2 ) s. (b) Time(s) when the particle is at rest. (c) Displacement and distance traveled from ( t = 0 ) to ( t = 5 ) s.

Solution (Mathalino structure):

Step 1: Find v(t) and a(t) [ v(t) = \fracdsdt = 3t^2 - 12t + 9 \quad (\textm/s) ] [ a(t) = \fracdvdt = 6t - 12 \quad (\textm/s^2) ]

Step 2: Evaluate at t=2 s [ v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3 \ \textm/s ] [ a(2) = 6(2) - 12 = 0 \ \textm/s^2 ]

Step 3: Particle at rest → ( v(t)=0 ) [ 3t^2 - 12t + 9 = 0 \implies t^2 - 4t + 3 = 0 \implies (t-1)(t-3)=0 ] Thus, ( t = 1 ) s and ( t = 3 ) s. By the time the long exam arrived, Miguel

Step 4: Displacement from t=0 to t=5 Compute positions: [ s(0) = 2,\ s(1) = 1 - 6 + 9 + 2 = 6,\ s(3) = 27 - 54 + 27 + 2 = 2,\ s(5) = 125 - 150 + 45 + 2 = 22 ] Displacement = ( s(5) - s(0) = 22 - 2 = 20 ) m (positive, to the right).

Step 5: Distance traveled – Need to account for direction changes at t=1 and t=3. From t=0 to 1: ( |s(1)-s(0)| = |6-2| = 4 ) m. From t=1 to 3: ( |s(3)-s(1)| = |2-6| = 4 ) m. From t=3 to 5: ( |s(5)-s(3)| = |22-2| = 20 ) m. Total distance = ( 4 + 4 + 20 = 28 ) m.

Answer: (a) ( v=-3 \ \textm/s, a=0 ); (b) ( t=1,3 \ \texts ); (c) Displacement = 20 m, Distance = 28 m.

| 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 |