Question

A ball moves in a straight line has an acceleration of a(t) = 2t + 5. Find the position function of the ball if its initial velocity is -3 cm per second and it’s initial position is 12 cm.

Answer 1

Answer:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$

Step-by-step explanation:

Relation between acceleration, velocity and position:

The velocity function is the integral of the acceleration function.

The position function is the integral of the velocity function.

Acceleration:

As given by the problem, the acceleration function is $a(t) = 2t + 5$

Velocity:

$v(t) = \int a(t) dt = \int (2t+5) dt = \frac{2t^2}{2} + 5t + K = t^2 + 5t + K$

In which K is the constant of integration, which is the initial velocity. So K = -3 and:

$v(t) = t^2 + 5t - 3$

Position:

$s(t) = \int v(t) dt = \int (t^2 + 5t - 3) = \frac{t^3}{3} + \frac{5t^2}{2} - 3t + K$

In which K, the constant of integration, is the initial position. Since it is 12:

$s(t) = frac{t^3}{3} + \frac{5t^2}{2} - 3t + 12$