a = 4 * t^2 – 2
At t = 0, a = -2 m/s^2
At t = 2, a = 4 * 2^2 – 2 = +14 m/s^2
At t = 4.3, a = 4 * 4.3^2 – 2 = 71.96 m/s^2
The acceleration is constantly increasing. The graph of acceleration vs time a parabola. The area under this graph is the velocity. The area under the velocity graph is the displacement. We have some integrating to do.
Velocity is the integral of acceleration
Integral of 4t^2 – 2 = 4/3t^3 – 2t + c
Position is the integral of velocity
Integral of 4/3t^3 – 2t + c = ⅓ * t^4 – t^2 + c * t + d
Position = ⅓ * t^4 – t^2 + c * t + d
Initial position = 1
At t = 0, position = 1
1 = ⅓ * 0^4 – 0^2 + c * 0 + d
1 = d
Position = ⅓ * t^4 – t^2 + c * t + 1
At t = 2, position = 20
20 = ⅓ * 2^4 – 2^2 + c * 2 + 1
20 = 5⅓ – 4 + c * 2 + 1
20 = 2⅓ + c * 2
17⅔ = c * 2
8.8333 = c
Position = ⅓ * t^4 – t^2 + 8.833 * t + 1
Position = ⅓ * 4.3^4 – 4.3^2 + 8.833 * 4.3 + 1
The answer is approximately 134.45 m