Example: Constant acceleration

Understanding the situation

Since accelerations are caused by unbalanced forces, we will have a lot more examples of calculations with acceleration once we have studied different forces and their properties. But to help us get a sense of what kind of problems one can solve with the constant acceleration equations, let's look at an example where the accelerations are specified directly (instead of specifying the forces, which is more natural).

Presenting a sample problem

By NASA/Wallops - <a class='external' href='http://sites.wff.nasa.gov/code840/multimedia.html'>http://sites.wff.nasa.gov/code840/multimedia.html</a> (direct link), Public Domain, <a class='external' href='https://commons.wikimedia.org/w/index.php?curid=464438'>https://commons.wikimedia.org/w/index.php?curid=464438</a>

Small rockets have been used to probe the atmosphere for 40 years. Scientific packages in the nose cone of the rocket detect chemicals in the atmosphere and return to earth by parachute. One early such rocket (the Canadian Black Brant VI) is shown in the figure at the right. It was quite small, being only about 9 ft long and weighing about 100 kg. Yet it could climb quite high. Let’s see if it could probe the stratosphere (10-50 km up), which contains the environmentally important ozone layer. Be sure to explain your reasoning for all parts of this problem. (For simplicity, take g for this problem to be 10 m/s2.)

A. When launched, the Black Brant VI could accelerate upward with an acceleration of 5g for 20 seconds. At the end of this acceleration phase, what was its velocity?

B. At the end of the acceleration phase, how high up would it be?

C. At the end of the acceleration phase, the rocket just turned off. It is now only under the influence of the force of gravity. If we ignore the effects of air resistance, it will continue to move with an acceleration of magnitude g pointing in the downward direction. Calculate the total height to which it will rise.

Solving this problem

Let's take our coordinate system so that the vertical direction is y and positive y is up. The foothold equations that are relevant for motion with a constant acceleration are

$$\langle \overrightarrow{v} \rangle = \frac{\Delta\overrightarrow{r}}{\Delta t}  \;\;\;\;\;\;\;\;\langle \overrightarrow{a} \rangle = \frac{\Delta\overrightarrow{v}}{\Delta t} $$

A. If the rocket accelerates upward with an acceleration = 5g, where g = 10 m/s2 for a period of 20 s, we can figure out the final velocity of this phase from the acceleration equation (call it $v_1$):

$$\langle \overrightarrow{a} \rangle = \frac{\Delta\overrightarrow{v}}{\Delta t} = \frac{v_f - v_i}{\Delta t} = \frac{v_1 - 0}{\Delta t} = \frac{v_1}{\Delta t}$$

B. We can get the height it rises in the first acceleration phase from the velocity equation. Since we are assuming uniform acceleration we can use that the average velocity is the ½ the sum of the initial and final velocities. We know the initial velocity (it's 0) and the final velocity (1000 m/s) so we can get the average velocity is

$\langle \overrightarrow{v} \rangle = \frac{v_i + v_f}{2} = \frac{0 + 1000 m/s}{2} = 500$ m/s

Solving the basic velocity equation for the change in position and using that it travelled at that acceleration for 20 s, we get

$\Delta y = \langle v \rangle \Delta t$ = ( 500 m/s)(20 s) = 10,000 m = 10 km.

C. When the engine turns off, the rocket is going upward with a speed of 1000 m/s. Even though it now has a downward acceleration, it will keep going up but slow down as a result of gravity. Each second, the force of gravity will subtract 10 m/s from the rocket's speed. It will reach its highest point (the top) when its velocity is 0. After that, subtracting 10 m/s each second will create a negative velocity and it will start to fall. We can calculate how high it went if we know the time it traveled and its average velocity. We know it went up 10 km in the first phase of acceleration (during which a = 5 m/s2). After the rocket turns off, let's see how far up it went in the second phase (during which a = -10 m/s2). We can use the acceleration equation to get the time and then the velocity equation to get the distance.

The time it takes to get to its highest point is given by the foothold acceleration equation which can easily be solved for the time (and you can probably do it in your head by just asking, "If I subtract 10 m/s from 1000 m/s, how many seconds will it be before my velocity is equal to 0?"). 

$$\langle a \rangle = \frac{\Delta v}{\Delta t} = \frac{v_f - v_i}{\Delta t} = \frac{0 - (1000\;m/s)}{\Delta t} = -\frac{1000\;m/s}{\Delta t}  $$

$$\Delta t = \frac{-1000\;m/s}{\langle a \rangle} = \frac{-1000\;m/s}{-10\;m/s^2}= 100\;s$$

So it will keep going up for another 100 seconds. We get the height by using the average velocity equation. For this phase of the motion it starts with a velocity of 1000 m/s and stops going up when the velocity get down to 0 m/s and it starts to fall back down. So the extra height (change in y) is

$$\langle v \rangle = \frac{\Delta y}{\Delta t} $$

$$\Delta y = \langle v \rangle \Delta t = \frac{v_i + v_f}{2} \Delta t = (\frac{1000\;m/s+0\;m/s}{2})(100\;s)$$

$\Delta y$ = (500 m/s)(100 s) = 50,000 m = 50 km

So it went up 10 km in the first phase and 50 km in the second for a total of 60 km. 

It's interesting to note that even though it had a downward acceleration in the second phase, it had an upward velocity so it kept rising. And since the upward acceleration was 5 times as great as the downward one, the time it was going up and slowing was 5 times as long as the time it was going up and speeding up -- and the distances traveled up were in the same proportion. Most of the upward travel took place after the rocket was turned off!

Joe Redish 8/3/15

Article 333
Last Modified: February 24, 2019