Question

answer the following questions. make sure to show all work to receive credit. you may need a separate sheet of paper. practice: acceleration math 1. a roller - coaster car rapidly picks up speed as it rolls down a slope. as it starts down the slope, its speed is 4 m/s. 3 seconds later, at the bottom of the slope, its speed is 22 m/s. find its acceleration. 2. a cyclist accelerates from 0 m/s to 8 m/s in 3 seconds. what is his acceleration? 3. a car approaches a stop sign going 5 m/s before it comes to a stop. if it does this in 4 seconds, what was its acceleration? a = (4 - 22)/3 a=-6 m/s² a=(0 - 8)/3 a=-2.7 m/s² a=(5 - 0)/4

Explanation:

Step1: Recall acceleration formula

The formula for acceleration is $a=\frac{v_f - v_i}{t}$, where $v_f$ is final - velocity, $v_i$ is initial - velocity, and $t$ is time.

Step2: Solve problem 1

Given $v_i = 22\ m/s$, $v_f = 4\ m/s$, and $t = 3\ s$. Substitute into the formula: $a=\frac{4 - 22}{3}=\frac{- 18}{3}=-6\ m/s^2$.

Step3: Solve problem 2

Given $v_i = 0\ m/s$, $v_f = 8\ m/s$, and $t = 3\ s$. Substitute into the formula: $a=\frac{8 - 0}{3}=\frac{8}{3}\approx2.67\ m/s^2$.

Step4: Solve problem 3

Given $v_i = 5\ m/s$, $v_f = 0\ m/s$, and $t = 4\ s$. Substitute into the formula: $a=\frac{0 - 5}{4}=-\frac{5}{4}=-1.25\ m/s^2$.

Answer:

1. $a=-6\ m/s^2$
2. $a=\frac{8}{3}\ m/s^2\approx2.67\ m/s^2$
3. $a =-\frac{5}{4}\ m/s^2=-1.25\ m/s^2$