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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?

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?

Grade:12th pass

1 Answers

Arun
25750 Points
6 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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