Students will learn the meaning of acceleration, how it is different than velocity and how to calculate average acceleration.
Key Equations
\begin{align*}v =\end{align*} velocity (m/s)
\begin{align*}v_i =\end{align*} initial velocity
\begin{align*}v_f =\end{align*} final velocity
\begin{align*}\Delta v =\end{align*} change in velocity \begin{align*}= v_f - v_i\end{align*}
\begin{align*}v_{avg} = \frac{\Delta x}{\Delta t}\end{align*}
\begin{align*}a =\end{align*} acceleration \begin{align*}(m/s^2)\end{align*}
\begin{align*}a_{avg} = \frac{\Delta v}{\Delta t}\end{align*}
- Acceleration is the rate of change of velocity. So in other words, acceleration tells you how quickly the velocity is increasing or decreasing. An acceleration of \begin{align*} 5 \ m/s^2 \end{align*} indicates that the velocity is increasing by \begin{align*} 5 m/s \end{align*} in the positive direction every second.
- Gravity near the Earth pulls an object downwards toward the surface of the Earth with an acceleration of \begin{align*}9.8 \ m/s^2 ( \approx 10 \ m/s^2)\end{align*}. In the absence of air resistance, all objects will fall with the same acceleration. The letter \begin{align*}g\end{align*}is used as the symbol for the acceleration of gravity.
- When talking about an object's acceleration, whether it is due to gravity or not, the acceleration of gravity is sometimes used as a unit of measurement where \begin{align*}1g=9.8m/s^2\end{align*}. So an object accelerating at 2g's is accelerating at \begin{align*}2*9.8m/s^2\end{align*} or \begin{align*}19.6m/s^2\end{align*}
- Deceleration is the term used when an object’s speed (i.e. magnitude of its velocity) is decreasing due to acceleration in the opposite direction of its velocity.
Example 1
A Top Fuel dragster can accelerate from 0 to 100 mph (160 km/hr) in 0.8 seconds. What is the average acceleration in \begin{align*}m/s^2\end{align*}?
Question: \begin{align*}a_{avg} = ? \ [m/s^2]\end{align*}
Given: \begin{align*}v_i = 0 \ m/s\end{align*}
\begin{align*}{\;} \qquad \ \ v_f = 160 \ km/hr\end{align*}
\begin{align*}{\;} \qquad \ \quad t = 0.8 \ s\end{align*}
Equation: \begin{align*}a_{avg} = \frac{\Delta v }{t}\end{align*}
Plug n’ Chug: Step 1: Convert km/hr to m/s
\begin{align*}v_f = \left( 160 \frac{km}{hr} \right ) \left( \frac{1,000 \ m}{1 \ km} \right ) \left ( \frac{1 \ hr}{3,600 \ s} \right ) = 44.4 \ m/s\end{align*}
Step 2: Solve for average acceleration:
\begin{align*}a_{avg} = \frac{\Delta v}{t} = \frac{v_f - v_i}{t} = \frac{44.4 \ m/s - 0 \ m/s}{0.8 \ s} = 56 \ m/s^2\end{align*}
Answer: \begin{align*}\boxed {\mathbf{56 \ m/s^2}}\end{align*} Note that this is over \begin{align*}5 \frac{1}{2}\end{align*} g’s!
Watch this Explanation
You can download this simultion and play around on it a bit to see the principles in action.
The Moving Man (PhET Simulation)
Time for Practice
- Ms. Reitman’s scooter starts from rest and accelerates at \begin{align*}2.0 m/s^2\end{align*}.
- Where will the scooter be relative to its starting point after 7.0 seconds?
- What is the scooter's velocity after 1s? after 2s? after 7s?
- Below are images from a race between Ashaan (above) and Zyan (below), two daring racecar drivers. High speed cameras took four pictures in rapid succession. The first picture shows the positions of the cars at \begin{align*}t = 0.0\end{align*}. Each car image to the right represents times 0.1, 0.2, and 0.3 seconds later.
- Who is ahead at \begin{align*}t = 0.2 \;\mathrm{s}\end{align*} ? Explain.
- Who is accelerating? Explain.
- Who is going fastest at \begin{align*} t = 0.3 \;\mathrm{s}\end{align*}? Explain.
- Which car has a constant velocity throughout? Explain.
- Graph \begin{align*}x\end{align*} vs. \begin{align*}t\end{align*} and \begin{align*}v\end{align*} vs. \begin{align*}t\end{align*}. Put both cars on same graph; label which line is which car.
- Which car is going faster at \begin{align*}t = 0.2 \;\mathrm{s}\end{align*} (Hint: Assume they travel the same distance between 0.1 and 0.2 seconds)?
Answers
1. a. 49 m b. 2 m/s, 4 m/s, 14 m/s
2. See Video above