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The relation between acceleration and time for the object is given by a=2t+4t^2. Calculate the position of the object at t=4sec?

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3 years ago

```							We need the position and velocity at t=0. Call these x₀ and v₀. They may be zero in your question but you haven't stated this. a = 2t + 4t² Since a = dv/dt: dv/dt = 2t + 4t² Integrate: v = t² + (4/3)t³ + c When t = 0, v = v₀ so: v₀ = 0² + (4/3)0³ + c c = v₀ v = t² + (4/3)t³ + v₀ Since v = dx/dt: dx/dt = t² + (4/3)t³ + v₀ Integrate: x = (1/3)t³ + (4/3)(1/4)t⁴ + v₀t + C . .= (1/3)t³ + (1/3)t⁴ + v₀t + C When t = 0, x = x₀ so: x₀ = (1/3)0³ + (1/3)0⁴ + v₀0 + C x₀ = C x = (1/3)t³ + (1/3)t⁴ + v₀t + x₀ When t=4s: x = (1/3)4³ + (1/3)4⁴ + 4v₀ + x₀ . = 64/3 + 256/3 + 4v₀ + x₀ If we take x₀ = 0 and v₀=0 this gives: x = (64 + 256)/3 x = 320/3 metres
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3 years ago
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