Rectilinear Motion Problems And Solutions Mathalino Upd

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.

When solving similar problems from Mathalino or recitation quizzes, avoid these mistakes:

Statement:
Velocity of a particle is ( v(t) = t^2 - 4t + 3 ) (m/s). Initial position ( s(0) = 0 ). Find:

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 ) rectilinear motion problems and solutions mathalino upd

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 )

Statement:
A particle moves with position ( s(t) = 2\sin(3t) ), ( t ) in seconds, ( s ) in meters. Find: Let t = time for first stone to hit ground

Solution:

1. ( v(t) = 6\cos(3t) )
( a(t) = -18\sin(3t) )

2. Max speed = max ( |v(t)| = |6\cos(3t)| = 6 , \textm/s ) (since max ( |\cos| = 1 )).

3. ( a(t) = 0 ) → ( -18\sin(3t) = 0 ) → ( \sin(3t) = 0 ) → ( 3t = n\pi ) → ( t = \fracn\pi3 )
Smallest positive ( t ): ( n=1 ) → ( t = \pi/3 \approx 1.047 , \texts )

Answers:
( v(t) = 6\cos(3t) ), ( a(t) = -18\sin(3t) )
Max speed = ( 6 , \textm/s )
First zero acceleration at ( t = \pi/3 , \texts )

| 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 | When solving similar problems from Mathalino or recitation

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.