An object starts from rest and has an acceleration given by a = 3.00 t^2 – 4.20 t. determine how far the object is from its starting point at the end of 4.50s using calculus.

38.7 units

Explanation:

a = 3.00 t² – 4.20 t

Integrate to find velocity as a function of time.

v = ∫ a dt

v = ∫ (3.00 t² – 4.20 t) dt

v = 1.00 t³ – 2.10 t² + C

The object starts at rest, so at t = 0, v = 0.

0 = 1.00 (0)³ – 2.10 (0)² + C

0 = C

v = 1.00 t³ – 2.10 t²

Integrate to get position as a function of time.

x = ∫ v dt

x = ∫ (1.00 t³ – 2.10 t²) dt

x = 0.250 t⁴ – 0.700 t³ + C

Find the difference in positions between t = 4.50 and t = 0.

Δx = [0.250 (4.50)⁴ – 0.700 (4.50)³ + C] – [0.250 (0)⁴ – 0.700 (0)³ + C]

Δx = 0.250 (4.50)⁴ – 0.700 (4.50)³

Δx = 38.7

The object moves 38.7 units.

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