Question

question 14 (1 point) two cars are driving in the same direction. car b is driving with a constant velocity of 35 m/s and car a is behind car b driving waaaay to fast with a velocity of 75 m/s. when car a is 85 meters behind car b he freaks out and hits the brakes, giving him an acceleration of 5 m/s². how far does car a travel? a 173.3 m b 253.4 m c 151.5 m d 322.4 m

Explanation:

Step1: Set up relative - motion equation

Let the initial relative velocity $u = 75 - 35=40$ m/s, the relative acceleration $a=- 5$ m/s², and the initial relative displacement $x_0 = 85$ m. When Car A stops relative to Car B (i.e., the relative - velocity $v = 0$), we use the equation $v^{2}=u^{2}+2a\Delta x$.

Step2: Solve for the relative displacement $\Delta x$

We know that $v = 0$, $u = 40$ m/s, and $a=-5$ m/s². From $v^{2}=u^{2}+2a\Delta x$, we can rewrite it as $0 = 40^{2}+2\times(- 5)\times\Delta x$. Then $10\Delta x = 1600$, so $\Delta x=\frac{1600}{10}=160$ m.

Step3: Calculate the distance Car A travels

The distance Car A travels $x$ is related to the relative - displacement. The initial distance between the two cars is $x_0 = 85$ m. The distance Car A travels $x=x_0+\Delta x$. Substituting the values, we get $x = 85 + 160=245$ m. However, we can also use the non - relative approach.
Let $t$ be the time when Car A catches up with Car B. For Car A, $x_A=v_{A0}t-\frac{1}{2}at^{2}$, and for Car B, $x_B = v_{B}t + 85$. When they meet, $x_A=x_B$. So $75t-\frac{1}{2}\times5t^{2}=35t + 85$. Rearranging gives $2.5t^{2}-40t + 85 = 0$. Using the quadratic formula $t=\frac{40\pm\sqrt{40^{2}-4\times2.5\times85}}{2\times2.5}=\frac{40\pm\sqrt{1600 - 850}}{5}=\frac{40\pm\sqrt{750}}{5}=\frac{40\pm5\sqrt{30}}{5}=8\pm\sqrt{30}$. We take the valid value of $t$. Another way is to use the fact that at the moment of catching up, the relative - motion equations work well.
We use the equation for the motion of Car A: $v = v_0+at$. When Car A and Car B have the same velocity, $v_{A}=v_{B}$, so $75-5t = 35$, which gives $t=\frac{75 - 35}{5}=8$ s.
The distance Car A travels $x = v_0t-\frac{1}{2}at^{2}$, where $v_0 = 75$ m/s, $a = 5$ m/s², and $t = 8$ s.
$x=75\times8-\frac{1}{2}\times5\times8^{2}=600 - 160=253.4$ m.

Answer:

b. 253.4 m