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

$F(t) = (16.8~N/s)~t$ $a(t) = \frac{F(t)}{m} = \frac{(16.8~N/s)~t}{45.0~kg}$ $a(t) = (0.3733~m/s^3)~t$ We can use $a(t)$ to find $v(t)$. $v(t) = v_0 + \int_{0}^{t}a(t)~dt$ $v(t) = 0 + \int_{0}^{t}(0.3733~m/s^3)~t~dt$ $v(t) = (0.1867~m/s^3)~t^2$ We can use $v(t)$ to find $x(t)$. $x(t) = x_0 + \int_{0}^{t}v(t)~dt$ $x(t) = 0 + \int_{0}^{t}(0.1867~m/s^3)~t^2~dt$ $x(t) = (0.06223~m/s^3)~t^3$ At t = 5.00 s: $x = (0.06223~m/s^3)(5.00~s)^3$ $x = 7.78~m$ The object travels 7.78 meters in the first 5.00 seconds.