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Class 12 Integrals · Short Answer2 Markskinematics_integration

Class 12 Maths · Integrals · Short Answer #57

An object has acceleration a(t)=cos(πt) m/s² and initial velocity v(0)=1/(2π) m/s. Find its net displacement during the first 1.5 seconds. Explain why this is not its total distance traveled.
Velocity during the first 1.5 seconds. t in seconds; v in m/s. Displacement is signed area.
Velocity during the first 1.5 seconds. t in seconds; v in m/s. Displacement is signed area.

✅ Step-by-Step Solution

Final Answer: Net displacement 3/(4π)+1/π² m; total distance is larger.
  1. Integrating acceleration and using v(0) gives v(t)=(1/π)[1/2+sin(πt)].
  2. A displacement primitive is F(t)=t/(2π)−cos(πt)/π². Therefore F(3/2)−F(0)=3/(4π)+1/π² m. The velocity becomes negative after t=7/6, so total distance requires ∫|v|dt instead.
  3. The net displacement is 3/(4π)+1/π² m. For comparison, total distance is [F(7/6)−F(0)]+[F(7/6)−F(3/2)]=5/(12π)+(1+√3)/π² m.

📋 CBSE Stepwise Marking Scheme

  • [0.5 marks] Integrating acceleration and using v(0) gives v(t)=(1/π)[1/2+sin(πt)].
  • [1 mark] A displacement primitive is F(t)=t/(2π)−cos(πt)/π². Therefore F(3/2)−F(0)=3/(4π)+1/π² m. The velocity becomes negative after t=7/6, so total distance requires ∫|v|dt instead.
  • [0.5 marks] The net displacement is 3/(4π)+1/π² m. For comparison, total distance is [F(7/6)−F(0)]+[F(7/6)−F(3/2)]=5/(12π)+(1+√3)/π² m.

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