Answer
a) $100\;\rm m$
b) See the graph below.
Work Step by Step
a)
We know that the driver will take 0.5 s to apply the brakes which is the reaction time. At this small time interval, the car is still moving at a constant speed of 30 m/s.
Then the car will move the rest of the 60 m to stop which means that the final velocity is zero.
And to find the minimum distance to stop the car from 40 m/s, we need to find the acceleration of the car.
And to find the acceleration of the car, we need to figure out the distance traveled during the reaction time.
$$\Delta x=v_xt=30\cdot 0.5=\bf 15\;\rm m$$
This means that the car will stop after $\Delta x=60-15=\color{blue}{\bf45}\;\rm m$ of applying brakes.
Now we have the distance traveled by the car when applying brakes, and we need to find the acceleration.
We can use the kinematic formula of velocity squared which is
$$v_{fx}^2=v_{ix}^2+2a_x\Delta x$$
Solving for $a_x$;
$$a_x=\dfrac{v_{fx}^2-v_{ix}^2}{2\Delta x}=\dfrac{0^2-30^3}{2\cdot 45}$$
Thus,
$$a_x=\color{blue}{\bf -10}\;\rm m/s^2$$
Now we need to find the total distance traveled by the car when the driver applies the brakes at 40 m/s.
First, we need to find the distance traveled during the reaction time.
$$x_1=40\cdot 0.5=\bf 20\;\rm m\tag 1$$
Now we can use the kinematic formula of velocity squared and solve for $\Delta x$;
$$\Delta x=\dfrac{v_{fx}^2-v_{ix}^2}{2a_x}=\dfrac{0^2-40^2}{2\cdot (-10)}=\bf 80\;\rm m\tag 2$$
Thus, from (1) and (2), the total distance needed to stop the same car from 40 m/s is
$$\Delta x_{tot}=x_1+\Delta x=20+80=\color{red}{\bf 100}\;\rm m$$
b) The position-versus-time graph for the motion of the car for the first case.
