Question

a particle is moving along a straight line with the acceleration a= (15 t− 7t^2/3) ft/s^2, where t? is in seconds. determine the velocity and the position of the particle as a function of time when t= 0, v= 0 and x= 15 ft.

Answer 1

Velocity:

$$v(t)=15t^2/2-7t^3/9,\\ x(t)=5t^3/2-7t^4/36.\\ v(0)=0,\\ x(0)=0,\\ t=9.4\text{ or } 1.5\text{ s for }x=15\text{ ft}.$$