Answer
The magnitude of acceleration is $~~6.23~m/s^2$
Work Step by Step
We can convert $80.5~km/h$ to units of $m/s$:
$(80.5~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 22.36~m/s$
Let $T$ be the reaction time.
The distance traveled before braking is $x_1 = 22.36~T$
Let $a$ be the deceleration during braking. We can find the distance $x_2$ the car travels during braking:
$v^2 = v_0^2+2ax_2$
$0 = v_0^2+2ax_2$
$x_2 = -\frac{v_0^2}{2a}$
$x_2 = -\frac{(22.36)^2}{2a}$
$x_2 = -\frac{250}{a}$
We can write an equation for the total distance during the two phases:
$x_1+x_2 = 56.7$
$22.36~T-\frac{250}{a} = 56.7$
We can convert $48.3~km/h$ to units of $m/s$:
$(48.3~km/h)\times (\frac{1000~m}{1~km})\times (\frac{1~h}{3600~s}) = 13.42~m/s$
Let $T$ be the reaction time.
The distance traveled before braking is $x_1 = 13.42~T$
Let $a$ be the deceleration during braking. We can find the distance $x_2$ the car travels during braking:
$v^2 = v_0^2+2ax_2$
$0 = v_0^2+2ax_2$
$x_2 = -\frac{v_0^2}{2a}$
$x_2 = -\frac{(13.42)^2}{2a}$
$x_2 = -\frac{90}{a}$
We can write an equation for the total distance during the two phases:
$x_1+x_2 = 56.7$
$13.42~T-\frac{90}{a} = 24.4$
$(-\frac{250}{90})\times(13.42~T-\frac{90}{a}) = (-\frac{250}{90})(24.4)$
$-37.28~T+\frac{250}{a} = -67.78$
We can add the two equations for the total distance in each case:
$-14.92~T = -11.08$
$T = \frac{11.08}{14.92}$
$T = 0.74~s$
We can find the acceleration:
$22.36~T-\frac{250}{a} = 56.7$
$\frac{250}{a} = 22.36~T-56.7$
$a = \frac{250}{22.36~T-56.7}$
$a = \frac{250}{(22.36)(0.74)-(56.7)}$
$a = -6.23~m/s^2$
The magnitude of acceleration is $~~6.23~m/s^2$
