Rectilinear Motion Problems And Solutions Mathalino Upd __full__

Rectilinear Motion Problems and Solutions (Updated)

Last updated: April 12, 2026
Tag: Dynamics, Engineering Mechanics, Calculus

7. Summary Table of Key Formulas

| Quantity | Variable a(t) | Constant a | |----------|---------------|-------------| | v(t) | ∫ a dt + C | v₀ + a t | | s(t) | ∫ v dt + C | s₀ + v₀ t + ½ a t² | | v² relationship | Not directly | v² = v₀² + 2a(s-s₀) |

Key Formulas Used

| Quantity | Definition | Unit (SI) | | --- | --- | --- | | Position | ( s(t) ) | m | | Velocity | ( v(t) = s'(t) ) | m/s | | Acceleration | ( a(t) = v'(t) = s''(t) ) | m/s² | | Constant acceleration | ( v = u + at ) | — | | | ( s = ut + \frac12 at^2 ) | — |

3. Problem Set #2: Constant Acceleration (Free-Fall)

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)

Problem 2: Variable Acceleration (Integration Required)

Statement: The acceleration of a particle in rectilinear motion is given by ( a(t) = 6t + 4 \ \textm/s^2 ). At ( t=0 ), the velocity ( v_0 = 5 \ \textm/s ) and position ( s_0 = 2 \ \textm ). Find the position function ( s(t) ). rectilinear motion problems and solutions mathalino upd

Solution:

Given ( a(t) = \fracdvdt = 6t + 4 ). Integrate: [ v(t) = \int (6t + 4) , dt = 3t^2 + 4t + C_1 ] Using ( v(0)=5 ): ( 5 = 0 + 0 + C_1 \implies C_1 = 5 ). Thus, ( v(t) = 3t^2 + 4t + 5 ).

Now, ( v(t) = \fracdsdt \implies s(t) = \int (3t^2 + 4t + 5) , dt = t^3 + 2t^2 + 5t + C_2 ). Using ( s(0)=2 ): ( 2 = 0 + 0 + 0 + C_2 \implies C_2 = 2 ).

Therefore, ( s(t) = t^3 + 2t^2 + 5t + 2 ) meters. Key Formulas Used | Quantity | Definition |

Answer: ( s(t) = t^3 + 2t^2 + 5t + 2 ).

Problem 3: Maximum Velocity & Position

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:

1. Velocity function ( v(t) )
2. Position function ( s(t) )
3. Time when velocity is maximum (if any)
4. Maximum velocity

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 ) ( a'(t) = 12 &gt

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.