Rectilinear Motion Problems And Solutions Mathalino Upd Direct

Used when the problem presents a graph (Velocity vs. Time).

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 rectilinear motion problems and solutions mathalino upd

✅ Answer: Car B overtakes after 24.14 seconds.

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: Used when the problem presents a graph (Velocity vs

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 ) Let s=0 at Car B’s initial position

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.