## University Physics with Modern Physics (14th Edition)

$a_x(t) = -(0.0320~m/s^3)(15.0~s-t)$ $v_x(t) = v_{0x}+ \int_{0}^{t}a_x(t)~dt$ $v_x(t) = v_{0x}+ \int_{0}^{t}-(0.0320~m/s^3)(15.0~s-t)~dt$ $v_x(t) = v_{0x}+ \int_{0}^{t}(0.0320~m/s^3)(t) - (0.480~m/s^2)~dt$ $v_x(t) = v_{0x}+ (0.0160~m/s^3)(t^2) - (0.480~m/s^2)(t)$ We know that $v_{0x} = 8.00~m/s$ at t=0. $v_{0x}+ (0.0160~m/s^3)(0^2) - (0.480~m/s^2)(0) = 8.00~m/s$ $v_{0x} = 8.00~m/s$ We can use $v_x(t)$ to find $x(t)$. $x(t) = x_0+ \int_{0}^{t}v_x(t)~dt$ $x(t) = x_0+ \int_{0}^{t}(0.0160~m/s^3)~t^2 - (0.480~m/s^2)~t+(8.00~m/s)~dt$ $x(t) = x_0 + \frac{0.0160~m/s^3}{3}~t^3 - (0.240~m/s^2)~t^2+(8.00~m/s)~t$ We know that x = -14.0 m when t=0. $x(t) = x_0 + \frac{0.0160~m/s^3}{3}~(0)^3 - (0.240~m/s^2)~(0)^2+(8.00~m/s)~(0) = -14.0~m$ $x_0 = -14.0~m$ $x(t) = \frac{0.0160~m/s^3}{3}~t^3 - (0.240~m/s^2)~t^2+(8.00~m/s)~t-14.0~m$ At t = 10.0 s, $x(t) = \frac{0.0160~m/s^3}{3}~(10.0~s)^3 - (0.240~m/s^2)~(10.0~s)^2+(8.00~m/s)~(10.0~s)-14.0~m$ $x(t) = 47.3~m$ The x-coordinate of the object when t = 10.0 s is 47.3 m.