Applied Mathematics — Second: Dynamics · Mathematics. These are 10 questions from the Magmo3 bank on this lesson, from easy to hard. Solve each one, then open "Show answer and explanation". Magmo3 has 74 questions on this lesson, each with an explanation.
A particle has velocity v = 4t + 1 (m/sec). Find its acceleration.
Correct answer: (B) 4 m/sec²
Explanation: Step 1: Acceleration is the derivative of velocity: a = dv/dt. Step 2: a = d/dt(4t + 1) = 4. Step 3: The acceleration is constant. ∴ acceleration = 4 m/sec²
The displacement of a body is s = t² − 6t + 5 (m). Find the time at which it is at rest.
Correct answer: (D) t = 3 s
Explanation: Step 1: v = ds/dt = 2t − 6. Step 2: At rest v = 0: 2t − 6 = 0. Step 3: 2t = 6, so t = 3. ∴ the body is at rest at t = 3 s
A particle has velocity v = 2t − 8 (m/sec). Find the time at which it comes to rest.
Correct answer: (C) t = 4 s
Explanation: Step 1: The body is at rest when v = 0: 2t − 8 = 0. Step 2: 2t = 8, so t = 4. ∴ the body is at rest at t = 4 s
A body moves so that s = 2t³ − 9t² + 12t (m). Find its acceleration at the first instant it comes to rest.
Correct answer: (B) −6 m/sec²
Explanation: Step 1: v = 6t² − 18t + 12 = 6(t − 1)(t − 2); first rest at t = 1. Step 2: a = dv/dt = 12t − 18. Step 3: At t = 1: a = 12 − 18 = −6. ∴ acceleration = −6 m/sec²
A particle moves so that s = t³ − 12t (m). Find the distance travelled between t = 0 and t = 4 s.
Correct answer: (B) 48 m
Explanation: Step 1: v = 3t² − 12 = 0 at t = 2 s (within 0 to 4). Step 2: s(0) = 0, s(2) = 8 − 24 = −16, s(4) = 64 − 48 = 16. Step 3: Distance 0 to 2 = |−16| = 16 m; distance 2 to 4 = |16 − (−16)| = 32 m. Step 4: Total = 16 + 32 = 48. ∴ distance travelled = 48 m
A body moves so that s = t⁴ − 4t³ + 10 (m). Find its acceleration when t = 2 s.
Correct answer: (C) 0 m/sec²
Explanation: Step 1: v = ds/dt = 4t³ − 12t². Step 2: a = dv/dt = 12t² − 24t. Step 3: At t = 2: a = 12(4) − 24(2) = 48 − 48 = 0. ∴ acceleration = 0 m/sec²
A body starts from rest at the origin with acceleration a = 6t + 2 (m/sec²). Find its displacement when t = 2 s.
Correct answer: (A) 12 m
Explanation: Step 1: v = ∫(6t + 2) dt = 3t² + 2t + c₁; v = 0 at t = 0 gives c₁ = 0. Step 2: s = ∫(3t² + 2t) dt = t³ + t² + c₂; s = 0 at t = 0 gives c₂ = 0. Step 3: At t = 2: s = 8 + 4 = 12. ∴ displacement = 12 m
A particle starts from a fixed point O on a straight line with velocity 3 m/sec. Its acceleration is a = 18e^(−2r) m/sec², where r is its position from O in metres. As r increases, the velocity of the particle approaches a limiting value. Find this value.
Correct answer: (A) 3√3 m/sec
Explanation: Step 1: a = v dv/dr, so ∫₃ᵛ v dv = ∫₀ʳ 18e^(−2r) dr. Step 2: ½(v² − 9) = [−9e^(−2r)]₀ʳ = 9(1 − e^(−2r)), so v² = 27 − 18e^(−2r). Step 3: As r increases, e^(−2r) gets closer and closer to 0, so v² increases towards 27. Step 4: The limiting velocity = √27 = 3√3 m/sec.
A particle has position vector r = t² î + (t² − 2t) ĵ (m). Find the time at which its speed is minimum.
Correct answer: (C) t = 0.5 s
Explanation: Step 1: v = 2t î + (2t − 2) ĵ, so ‖v‖² = (2t)² + (2t − 2)² = 8t² − 8t + 4. Step 2: Speed is least when d(‖v‖²)/dt = 0: 16t − 8 = 0. Step 3: t = 0.5. ∴ time = 0.5 s
A particle moves in a straight line with acceleration a = 2x (m/sec²), where x is its displacement in metres. If its velocity is 2 m/sec when x = 0, find its velocity when x = √6 m.
Correct answer: (A) 4 m/sec
Explanation: Step 1: a = v dv/dx ⇒ v dv = 2x dx ⇒ v²/2 = x² + c. Step 2: v = 2 at x = 0 ⇒ c = 2, so v² = 2x² + 4. ∴ At x = √6: v² = 12 + 4 = 16, v = 4 m/sec.
The first 10 questions in every subject are free, each with an explanation.