Question

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. what is its acceleration? a = (vf - vi)/t, vi = 4m/s, vf = 22m/s, t = 3s, a=(22 - 4)/3, a = 6m/s². 2. a cyclist slows down from 8 m/s to a stop in 3 seconds. what is his acceleration? a=(vf - vi)/t, vf = 0m/s, vi = 8m/s, t = 3s, a=(0 - 8)/3, a=-8/3m/s², a≈ - 2.7m/s². 3. a car approaches a stop sign going 5 m/s. if it does this in 4 seconds, what was its acceleration? a=(vf - vi)/t, vf = 0m/s, vi = 5m/s, t = 4s, a=(0 - 5)/4, a=-5/4m/s². 4. a car advertisement claims that a certain car can accelerate from rest to 70 km/hr in 7 seconds (hint: convert to m/s first!). find the cars acceleration. 5. challenge: if a ferrari with an initial velocity of 10 m/s accelerates at a rate of 50 m/s² for 3 seconds, what will its final velocity be?

Explanation:

Step1: Identify the 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: Analyze problem 4

Given $v_i = 0$ (starts from rest), $v_f=70\ km/h$. First convert $70\ km/h$ to $m/s$. $70\ km/h=\frac{70\times1000}{3600}\ m/s\approx 19.44\ m/s$, and $t = 7\ s$. Then $a=\frac{v_f - v_i}{t}=\frac{19.44 - 0}{7}\ m/s^2\approx 2.78\ m/s^2$.

Step3: Analyze problem 5

Given $v_i$ (initial velocity), $a$ (acceleration) and $t$ (time), and we use the formula $v_f=v_i+at$. Assume $v_i = 0$ (if not given otherwise in the problem - context for the Ferrari starting situation), $a = 10\ m/s^2$ and $t = 4\ s$. Then $v_f=0 + 10\times4=40\ m/s$.

Answer:

For problem 4: The car's acceleration is approximately $2.78\ m/s^2$.
For problem 5: The final velocity of the Ferrari is $40\ m/s$.