Answer
$a(1) = 4.59$ ft s$^{-2}$
$a(5) = 1$ ft s$^{-2}$
$a(10) = 0.36$ ft s$^{-2}$
Work Step by Step
1. Find the derivative of $v(t)$ to get $a(t)$, the acceleration function
$v(t) = \frac{90t}{4t+10}$
$v'(t)/a(t) = \frac{(90)(4t+10)-(4)(90t)}{(4t+10)^{2}}$
$a(t) = \frac{360t+900-360t}{(4t+10)^{2}}$
$a(t) = \frac{900}{(4t+10)^{2}}$
2. Find the acceleration when $t = 1$
$a(1) = \frac{900}{(4(1)+10)^{2}}$
$a(1) = \frac{900}{(4+10)^{2}}$
$a(1) = \frac{900}{(14)^{2}}$
$a(1) = \frac{900}{196}$
by GDC / calculator
$a(1) = 4.59$ ft s$^{-2}$
2. Find the acceleration when $t = 5$
$a(5) = \frac{900}{(4(5)+10)^{2}}$
$a(5) = \frac{900}{(20+10)^{2}}$
$a(5) = \frac{900}{(30)^{2}}$
$a(5) = \frac{900}{900}$
by GDC / calculator
$a(5) = 1$ ft s$^{-2}$
3. Find the acceleration when t = 10
$a(10) = \frac{900}{(4(10)+10)^{2}}$
$a(10) = \frac{900}{(40+10)^{2}}$
$a(10) = \frac{900}{(50)^{2}}$
$a(10) = \frac{900}{2500}$
by GDC / calculator
$a(10) = 0.36$ ft s$^{-2}$
