Answer to Question #267141 in Classical Mechanics for Rex

Question

a particle is moving along a straight line with the acceleration a= (15 t− 7t/3) ft/s2, 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

$a(t) = 15t-\frac{7}{3}t = \frac{38}{3}t$

$v(t) = \int a(t)dt= \frac{38}{6}t^2+C_1$

$v(0) = 0 ; C_1=0$

$v(t) = 6t^2+\frac{t^2}{3}$

$x(t)= \int v(t)dt= 2t^3+\frac{t^3}{9}+C_2$

$x(0) =15; C_2 = 15$

$\text{Answer:}$

$v(t) = 6t^2+\frac{t^2}{3}$

$x(t)= \int v(t)dt= 2t^3+\frac{t^3}{9}+15$