Answer
(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) $F_{thrust} = 28,700~N$
Work Step by Step
(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.
$\sum F = ma$
$F_{thrust} - mg = ma$
$F_{thrust} = ma+mg$
$F_{thrust} = (2540~kg)(1.50~m/s^2)+(2540~kg)(9.80~m/s^2)$
$F_{thrust} = 28,700~N$
