a car is moving with constant speed $v_0$ when a tree falls into the roadway at $t = 0$. it takes the driver…

a car is moving with constant speed $v_0$ when a tree falls into the roadway at $t = 0$. it takes the driver a time $t$ to react and apply the brakes. the car then slows with a constant acceleration of magnitude $a$. which of the following equations correctly expresses the distance $d$ traveled by the car from time $t = 0$ to the instant the car comes to rest?\n(a) $d = v_0t+\frac{1}{2}at^2$\n(b) $d=\frac{1}{2}at^2+\frac{v_0^2}{2a}$\n(c) $d = v_0t+\frac{v_0^2}{2a}$\n(d) $d = v_0t+\frac{1}{2}at^2+\frac{v_0^2}{2a}$

a car is moving with constant speed $v_0$ when a tree falls into the roadway at $t = 0$. it takes the driver a time $t$ to react and apply the brakes. the car then slows with a constant acceleration of magnitude $a$. which of the following equations correctly expresses the distance $d$ traveled by the car from time $t = 0$ to the instant the car comes to rest?\n(a) $d = v_0t+\frac{1}{2}at^2$\n(b) $d=\frac{1}{2}at^2+\frac{v_0^2}{2a}$\n(c) $d = v_0t+\frac{v_0^2}{2a}$\n(d) $d = v_0t+\frac{1}{2}at^2+\frac{v_0^2}{2a}$

Answer

Explanation:

Step1: Calculate distance during reaction time

During the reaction - time (T), the car moves with a constant speed (v_0). Using the formula (d = vt), the distance (d_1) traveled during the reaction - time is (d_1=v_0T).

Step2: Calculate distance during braking

After applying the brakes, the car has an initial velocity (v_0), a final velocity (v = 0), and an acceleration (a). Using the kinematic equation (v^{2}-v_{0}^{2}=2ad), we can solve for the distance (d_2) during braking. Rearranging for (d) gives (d_2=\frac{v^{2}-v_{0}^{2}}{2a}). Since (v = 0), (d_2=\frac{0 - v_{0}^{2}}{-2a}=\frac{v_{0}^{2}}{2a}) (the acceleration is negative during braking, but we use the magnitude (a) in the formula).

Step3: Calculate total distance

The total distance (D) traveled by the car from (t = 0) to when it stops is the sum of the distance during the reaction - time and the distance during braking. So (D=d_1 + d_2=v_0T+\frac{v_{0}^{2}}{2a}).

Answer:

C. (D = v_0T+\frac{v_{0}^{2}}{2a})