The body moves in a straight line, the distance traveled s from time t is determined by the equation s (t) = – 2t + 2t ^ 3

The body moves in a straight line, the distance traveled s from time t is determined by the equation s (t) = – 2t + 2t ^ 3 to find the average speed of movement on the interval [2; 3] instantaneous speed and acceleration.

General solution.

We transform the given equation of motion:

s (t) = 2 * t ^ 3 – 2 * t = 2 * t * (t² – 1), that is, s (t) = 2 * t * (t² – 1).

The average speed V (t) is the distance s (t) divided by the time t, that is:

V (t) = s (t) / t = 2 * t * (t² – 1) / t = 2 * (t² – 1).

For our equation of motion:

Acceleration а = (Vк – Vн) / (tк – tн) = (2 * (tк² – 1) – 2 * (tн² – 1)) / (tк – tн) =

= 2 / (tк – tн) * (tк² – 2 – tн² + 2) = (tк² – tн²) * 2 / (tк – tн) =

= (tк – tн) * (tк + tн) * 2 / (tк – tн) = 2 * (tк + tн).

that is, a = 2 * (tk + tn).

The instantaneous speed Vm is the small difference between the initial s (tн) and the final distance s (tк), divided by a very small time interval during which this distance has changed:

Vm = (2 * tк * (tк² – 1) – 2 * tн * (tн² – 1)) / (tк – tн).

For a given interval t [2; 3], substitute the numbers:

The average speed for time t = 3 will be: V (3) = 2 * (3² – 1) = 16 (m / s);

and for time t = 2: V (2) = 2 * (2² – 1) = 6 (m / s);

Acceleration а = 2 * (tк + tн) = 2 * (3 + 2) = 10 (m / s²).

Instantaneous speed: Vm = (2 * 3 * (3² – 1) – 2 * 2 * (2² – 1)) / (3 – 2) =

= 6 * 8 – 4 * 3 = 48 – 12 = 36 (m / s).