Answer
Magnitude of braking acceleration $= 0.766\space m/s^{2}$
Work Step by Step
Please see the attached image first.
First of all, let's convert 45 km/h into m/s.
We know, 1 km = 1000 m and we can multiply the given velocity by 1000 m/km, also 1 h = 3600 s, then we can use a 3600 s/h conversion factor and gives,
$U=45\space km/h = (\frac{45\space km}{h})(\frac{1000\space m}{km})(\frac{h}{3600\space s})$
$U=\frac{50}{4}\space m/s$
$U=12.5\space m/s$
Let's apply the equation $V^{2}=U^{2}+2aS$ to find the acceleration of the vehicle.
$\rightarrow\space V^{2}=U^{2}+2aS$
Now plug the known values into this equation.
$\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space\space0^{2}\space\space\space\space\space\space=(12.5\space m/s)^{2}+ 2a\times102\space m$
$-156.25\space m^{2}/s^{2}= 204a\space m$
$\space\space\space\space\frac{-156.25}{204}\space m/s^{2}=a$
$\space\space\space\space \space\space\space\space \space\space\space\space \space\space\space\space a\space\space\space\space\space\space\space\space=-0.766\space m/s^{2}$
Magnitude of braking acceleration to avoid hitting the moose = $0.766\space m/s^{2}$
