Question

luego (mal7525) - unit 1 hw2 25 - 26 - tejeda - (barreralphy1_1) 3 already been released and is moving upward at time t = 0, turns around at t1, and hits the ground at t2. which of the following curves could describe the acceleration of the rock? 008 10.0 points an engineer in a locomotive sees a car stuck on the track at a railroad crossing in front of the train. when the engineer first sees the car, the locomotive is 320 m from the crossing and its speed is 21 m/s. if the engineers reaction time is 0.22 s, what should be the magnitude of the minimum deceleration to avoid an accident? answer in units of m/s². 009 10.0 points henry has tossed a rock upward. it has

Explanation:

Step1: Calculate distance during reaction - time

The train moves at a constant - speed during the reaction time. The distance $d_1$ covered during the reaction time $t_r = 0.22\ s$ with a speed $v = 21\ m/s$ is given by the formula $d_1=v\times t_r$.
$d_1 = 21\times0.22=4.62\ m$

Step2: Calculate the distance for deceleration

The distance available for deceleration $d_2$ is the total distance from the crossing minus the distance covered during the reaction time. The total distance from the crossing is $d = 320\ m$, so $d_2=d - d_1=320 - 4.62 = 315.38\ m$

Step3: Use the kinematic equation

The kinematic equation $v_f^2=v^2+2a\Delta x$, where the final velocity $v_f = 0\ m/s$, the initial velocity $v = 21\ m/s$, and the displacement $\Delta x=d_2 = 315.38\ m$.
We can re - arrange the equation $v_f^2=v^2+2a\Delta x$ for acceleration $a$. Since $v_f = 0$, we have $0 = v^2+2ad_2$, and $a=-\frac{v^2}{2d_2}$
Substitute $v = 21\ m/s$ and $d_2 = 315.38\ m$ into the formula:
$a=-\frac{21^2}{2\times315.38}=-\frac{441}{630.76}\approx - 0.7\ m/s^2$
The magnitude of the deceleration is $|a|\approx0.7\ m/s^2$

Answer:

$0.7\ m/s^2$