Step 1: Differentiate to get v(t) and a(t).
s(t) = 2t³ – 9t² + 12t + 5
v(t) = ds/dt = 6t² – 18t + 12
a(t) = dv/dt = 12t – 18
Step 2: Evaluate at t = 2 s.
v(2) = 6(4) – 18(2) + 12 = 24 – 36 + 12 = 0 m/s
a(2) = 12(2) – 18 = 24 – 18 = 6 m/s²
Step 3: Particle at rest means v(t) = 0.
6t² – 18t + 12 = 0 → divide 6: t² – 3t + 2 = 0 → (t-1)(t-2) = 0
Thus at t = 1 s and t = 2 s, the particle is momentarily at rest.
Step 4: Total distance traveled (t=0 to t=4).
We check the sign of velocity in intervals [0,1], [1,2], [2,4].
Compute positions:
s(0)=5 m
s(1)=2-9+12+5=10 m
s(2)=16-36+24+5=9 m
s(4)=128-144+48+5=37 m
Distance:
From 0→1: |10-5| = 5 m
From 1→2: |9-10| = 1 m
From 2→4: |37-9| = 28 m
Total distance = 5 + 1 + 28 = 34 m.
✅ Answer: (a) v=0, a=6 m/s²; (b) t=1 s, 2 s; (c) 34 m. rectilinear motion problems and solutions mathalino upd
Before diving into problems, recall the definitions:
A particle moves along a straight line (e.g., the x-axis). Its position at time ( t ) is given by ( s = f(t) ).
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 1: Differentiate to get v(t) and a(t)
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.
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 4:
Car A and Car B are on the same straight road. Car A is 100 m ahead of Car B at t=0. Car A moves with constant velocity 20 m/s. Car B starts from rest at t=0 and accelerates at 2 m/s². How long will it take for Car B to overtake Car A?
Problem: Car A starts from rest and accelerates at 2 m/s^2. How far in 5 s? Solution: s = 0 + 0·5 + 0.5·2·5^2 = 25 m. Before diving into problems, recall the definitions: A
a) Velocity: a = dv/dt = 4 - t² → dv = (4 - t²) dt
Integrate: v(t) = ∫(4 - t²) dt = 4t - t³/3 + C
At t=0, v=3 → 3 = 0 - 0 + C → C=3.
Thus v(t) = 4t - t³/3 + 3 m/s.
b) Position: v = ds/dt = 4t - t³/3 + 3 → ds = (4t - t³/3 + 3) dt
s(t) = ∫(4t - t³/3 + 3) dt = 2t² - t⁴/12 + 3t + D
At t=0, s=2 → 2 = 0 - 0 + 0 + D → D=2.
Thus s(t) = 2t² - t⁴/12 + 3t + 2 m.
c) v at t=4 s:
v(4) = 4(4) - (64)/3 + 3 = 16 - 21.333 + 3 = -2.333 m/s (moving negative direction).
d) Displacement from 0 to 4:
s(4) = 2(16) - (256)/12 + 3(4) + 2 = 32 - 21.333 + 12 + 2 = 24.667 m
s(0)=2 m → Displacement = 24.667 - 2 = 22.667 m.
✅ Answer: (a) v(t)=4t - t³/3+3; (b) s(t)=2t² - t⁴/12+3t+2; (c) -2.333 m/s; (d) 22.667 m.