Question

(02.02 mc)
a train is slowing down with an average acceleration of -6.0 m/s². if its initial velocity is 60.0 m/s, how far does it travel in 5.0 seconds?

a 140 m
b 230 m
c 270 m
d 530 m

Explanation:

Step1: Identify the kinematic - equation

We use the equation $x = v_0t+\frac{1}{2}at^{2}$, where $x$ is the displacement, $v_0$ is the initial velocity, $a$ is the acceleration, and $t$ is the time.

Step2: Substitute the given values

Given $v_0 = 60.0\ m/s$, $a=-6.0\ m/s^{2}$, and $t = 5.0\ s$.
Substitute into the equation: $x=(60.0\ m/s)\times5.0\ s+\frac{1}{2}\times(- 6.0\ m/s^{2})\times(5.0\ s)^{2}$.

Step3: Calculate each term

First term: $(60.0\ m/s)\times5.0\ s = 300\ m$.
Second term: $\frac{1}{2}\times(-6.0\ m/s^{2})\times(5.0\ s)^{2}=\frac{-6.0}{2}\times25\ m=-75\ m$.

Step4: Find the displacement

$x=300\ m - 75\ m=225\ m\approx230\ m$.

Answer:

B. 230 m