Question

a cart initially traveling at +10 m/s reaches the bottom of a hill. it accelerates up it at -2.0 m/s². what is the cart’s velocity 8.0 s after it starts accelerating? -6 questions 2 - 3 a ball released from rest accelerates down a ramp at 2 m/s². question 2 how far does it go in one second? 1 pts

Explanation:

Step1: Identify the kinematic - equation

We use the equation $x = v_0t+\frac{1}{2}at^{2}$. Since the ball is released from rest, $v_0 = 0\ m/s$, $a = 2\ m/s^{2}$, and $t = 1\ s$.

Step2: Substitute the values

Substitute $v_0 = 0$, $a = 2\ m/s^{2}$, and $t = 1\ s$ into the equation $x=v_0t + \frac{1}{2}at^{2}$. We get $x=0\times1+\frac{1}{2}\times2\times1^{2}$.

Step3: Calculate the result

$x=\frac{1}{2}\times2\times1 = 1\ m$.

Answer:

$1\ m$