91Ó°ÊÓ

Initial speed of the car, $u=20m/s$

Reaction time before striking the break, $t=0.50s.$

Acceleration of the car, role="math" localid="1648219673493" $a=-10m/s^2$.

Distance between the car and the deer,$d=35m.$

The distance travelled before pressing the break is,

$$\begin{gathered} s=ut \\ s=\left(20×0.50\right)m \\ s=10m \end{gathered}$$

Time duration of deacceleration of the car during stepping on the brakes can be determines as,

$$\begin{gathered} v=u+at\text{'} \\ 0=\left(20m/s\right)-\left(10m/s^2×t\text{'}\right) \\ t\text{'}=\frac{20}{10}s \\ t\text{'}=2s \end{gathered}$$

Distance covered during the deacceleration,

$$\begin{gathered} S=ut\text{'}+\frac{1}{2}at{\text{'}}^2 \\ S=\left(20×2\right)+\frac{1}{2}\left(-10\right)×2^2 \\ S=\left(40-20\right)m \\ S=20m \end{gathered}$$

Distance travelled during the uniform motion and deacceleration,$(s+S)=\left(10+20\right)m=30m.$

Before coming to a stop, the total distance from the deer is,

$\left(35-30\right)m=5m.$

Acceleration of the car, $a=-10m/s^2$.

Distance between the car and the deer, role="math" localid="1648219786110" $d=-35m.$

We have to find the maximum speed of the car without hitting the deer as $v_{max}$.

The total distance covered by the car from the starting time of its journey to stop without hitting the deer is,

$$\begin{gathered} d\text{'}=\left(S+5\right)m \\ d\text{'}=\left(20+5\right)m \\ d\text{'}=25m \end{gathered}$$

The maximum speed of the car should have, can be found as, $v_{max}^2=v_{final}^2-2ad\text{'}$

Now, $v_{final}=0$

Therefore,

localid="1650273158159" $$\begin{gathered} v_{max}^2=-2×10×25 \\ \left|v_{max}^2\right|=500 \\ {v}_{max}=\sqrt{500} \\ {v}_{max}=22.36m/s \end{gathered}$$