Answer
Decelerate at $21.5ft/s^2$
Work Step by Step
Step 1. Assuming we need a constant deceleration of $a=\frac{d^2s}{dt^2}=-k$, we can get $v=\frac{ds}{dt}=\int(-k)dt=-kt+C$ and $s=\int(-kt+C)dt=-\frac{k}{2}t^2+Ct+D$, where $C. D$ are constants.
Step 2. At $t=0, v=44ft/sec$, we have $-k(0)+C=44$ and $C=44$
Step 3. At $t=0, s=0$, we have $-\frac{k}{2}(0)^2+C(0)+D=0$ and $D=0$
Step 4. When the motorcycle stops, $v(t)=0$; thus $-kt+44=0$ and $t=\frac{44}{k} $
Step 5. For the above time $s(t)=45$, we have $-\frac{k}{2}(\frac{44}{k})^2+44(\frac{44}{k})=45$ which leads to $90k=44^2$ and $k=\frac{968}{45}\approx21.5ft/s^2$
