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

(a) $A = 1.50~m/s^2$ $B = 0.50~m/s^3$ (b) $a = 5.50~m/s^2$ (c) $F_{thrust} = 38,900~N$ $F_{thrust} = 1.56 \times ~w$ (d) $F_{thrust} = 28,700~N$
(a) $v(t) = At+Bt^2$ $a(t) = \frac{dv}{dt} = A+2Bt$ At t = 0, $a = 1.50~m/s^2$ $A + 2B(0) = 1.50~m/s^2$ $A = 1.50~m/s^2$ At t = 1.00 s, $v = 2.00~m/s$ $v = (1.50~m/s^2)(1.00~s)+B(1.00~s)^2 = 2.00~m/s$ $B = \frac{0.50~m/s}{1.00~s^2} = 0.50~m/s^3$ (b) At t = 4.00 s: $a = (1.50~m/s^2)+(2)(0.50~m/s^3)(4.00~s)$ $a = 5.50~m/s^2$ (c) At t = 4.00 s: $\sum F = ma$ $F_{thrust} - mg = ma$ $F_{thrust} = ma+mg$ $F_{thrust} = (2540~kg)(5.50~m/s^2)+(2540~kg)(9.80~m/s^2)$ $F_{thrust} = 38,900~N$ We can express the thrust as a multiple of the rocket's weight $w$ (which is equal to $mg$). $F_{thrust} = \frac{m(a+g)}{mg} = \frac{a+g}{g}$ $F_{thrust} = \frac{(5.50~m/s^2)+(9.80~m/s^2)}{9.80~m/s^2}$ $F_{thrust} = 1.56 \times ~w$ (d) At t = 0, $a = 1.50~m/s^2$ We can find the initial thrust. $\sum F = ma$ $F_{thrust} - mg = ma$ $F_{thrust} = ma+mg$ $F_{thrust} = (2540~kg)(1.50~m/s^2)+(2540~kg)(9.80~m/s^2)$ $F_{thrust} = 28,700~N$