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

$a_y(t) = (2.80~m/s^3)~t$ Note that $v_{0y} = 0$ and $y_0 = 0$ because the rocket is initially on the ground at rest. $v_y(t) = v_{0y}+ \int_{0}^{t}a_y(t)~dt$ $v_y(t) = 0 + \int_{0}^{t}(2.80~m/s^3)~t~dt$ $v_y(t) = (1.40~m/s^3)~t^2$ We can use $v_y(t)$ to find $y(t)$. $y(t) = y_0+ \int_{0}^{t}v_y(t)~dt$ $y(t) = 0+ \int_{0}^{t} (1.40~m/s^3)~t^2~dt$ $y(t) = \frac{1.40~m/s^3}{3}~t^3$ (a) At t = 10.0 s, $y = \frac{1.40~m/s^3}{3}~(10.0~s)^3 = 467~m$ At t = 10.0 s, the rocket is a height of 467 meters above the Earth's surface. (b) We can find the time $t$ when y = 325 m. $y(t) = \frac{1.40~m/s^3}{3}~t^3 = 325~m$ $t^3 = \frac{(325~m)(3)}{1.40~m/s^3}$ $t = 8.864~s$ At t = 8.864 s, $v_y = (1.40~m/s^3)~(8.864~s)^2$ $v_y = 110~m/s$ The speed is 110 m/s when the rocket is 325 meters above the Earth's surface.