64 Acceleration During a Skydive

Acceleration

After the becomes large enough to balance out a skydiver’s , they will have no . From we already know that an object’s prevents a change in unless it experience a net force, so from that point when the forces are balanced and onward, the skydiver continues at a until they open their parachute.

During the initial part of skydive, before the is large enough to balance out the weight,  there is a so their does change.  The rate at which the velocity changes is known as the . Note that students often confuse velocity and acceleration because they are both rates of change, so to be specific: velocity defines the rate at which the position is changing and acceleration defines the rate at which the velocity is changing.  We can calculate the (a) during a certain time interval (Δt) by subtracting the (vi) from the  (vf) to get the change in velocity (Δv) and then dividing by time interval (Δt):

(1)   \begin{equation*} \bold{a} = \frac{\bold{\Delta v}}{\Delta t} =\frac{\bold{v_f-v_i}}{\Delta t} \end{equation*}

Everyday Example

Let’s calculate the during the roughly 2 seconds it takes a parachute to fully open and slow a skydiver from  120 MPH to 6 MPH. First let’s remember that the skydiver is moving in our negative direction so the initial and final velocities should be negative. Also, lets convert to meters per second: \bold{v_f} = -2.7 \,\bold{m/s} and \bold{v_i}= -54 \,\bold{m/s}.

Starting with our definition of acceleration:

    \begin{equation*} \bold{a}  =\frac{\bold{v_f-v_i}}{\Delta t} \end{equation*}

Inserting our values:

    \begin{equation*} \bold{a} =\frac{-2.7 \,\bold{m/s}-(-54 \,\bold{m/s})}{2\,\bold{s})} \end{equation*}

The two negatives in front of the 54 m/s make a positive, and then we calculate a value.

    \begin{equation*} \bold{a} =\frac{-2.7 \,\bold{m/s}+(54 \,\bold{m/s})}{2\,\bold{s)}}   \approx 30 \,\bold{m/s/s} \end{equation*}

We now get a chance to see that the units of are m/s/s or equivalently m/s2

Acceleration Direction

The direction of depends on the direction of the change in . If the velocity becomes more negative, then acceleration must be negative. This is the case for our skydiver during the first part of the jump; their speed is increasing in the negative direction, so their velocity is becoming more negative, so their acceleration is negative. Conversely, if an object moves in the negative direction, but slows down, the acceleration is positive, even though the velocity is still negative! This was the case for our skydiver just after opening their parachute, when they still moved downward, but were slowing down. Slowing down in the negative direction means the velocity is becoming less negative, so the acceleration must be positive. All of the possible combinations of direction and change and the resulting are summarized in the following chart:

Table Showing Possible Acceleration Directions
Initial direction of motion ( initial velocity direction) Speed change Direction of Acceleration
positive speeding up positive
positive slowing down negative
negative speeding up negative
negative slowing down positive

Reinforcement Exercises

A car is moving in the positive direction and slams on the brakes. What direction is the ?

A car is moving in the negative direction and slams on the brakes. What direction is the acceleration?

A car is moving in the negative direction and speeding up. What direction is the acceleration?

Accelerating the Body

tells us that we need a in order to create an .   As you might expect, a larger net force will cause a larger acceleration, but the more matter you are trying to accelerate the larger force will be required. summarizes all of that into a single equation relating the , , and :

(2)   \begin{equation*} \bold{F_net} = m\bold{a} \end{equation*}

We can think of Newton’s Second Law in terms of the acceleration instead:

(3)   \begin{equation*} \bold{a} = \frac{\bold{F_net}}{m} \end{equation*}

Now we see that larger create larger and larger masses reduce the size of the acceleration. In fact, an object’s is a direct measure of an objects resistance to changing its motion, or its .

Check out this simulation to see how forces combine to create net forces and accelerations:

Forces and Motion: Basics

 

Reinforcement Exercises

You slide a box across the floor by applying a 220 N force to the right. applies a reactive 170 N force on the box to the left.

What is the size and direction of the on the box?

The box has a mass of 25 kg. What is the size and direction of the on the box?

The box is not accelerating in the vertical direction, so what is the net vertical force? [Hint: Forces in different dimensions (vertical and horizontal) don’t affect one another and applies separately to each dimension.]

How big is the on the box?

What is the value of the kinetic?

It’s an interesting quirk of our universe that the same property of an object, specifically its , determines both the on it and its resistance to accelerations, or . Said another way,  the inertial mass and the gravitational mass are equivalent. This result produces some interesting phenomena, as we will see in the following example.

Everyday Example

Let’s calculate the initial of our skydiver the moment they jump. At this moment they have the pulling them down, but they have not yet gained any , so the () is zero. The is then just gravity, because it is the only force. Starting with :

(4)   \begin{equation*} \bold{a} = \frac{\bold{F_{net}}}{m} \end{equation*}

And inserting our known formula for calculating force of gravity near the surface of Earth and including a negative sign because down is our negative direction, (\bold{F_g} = -mg):

(5)   \begin{equation*} \bold{a} = \frac{-mg}{m} \end{equation*}

We see that the cancels out,

(6)   \begin{equation*} \bold{a} = \frac{-\cancel{m}g}{\cancel{m}} = -g = -9.81 \bold{\frac{m}{s^2}} \end{equation*}

We see that our is negative, which makes sense because the acceleration is downward. We also see that the size of the acceleration is g = 9.81 m/s2. We have just shown that in the absence of , all objects falling near the surface of Earth will experience an  equal in size to g = 9.81 m/s2, regardless of their and .

In the previous example we found that our skydiver has an initial of g = 9.81 m/s2, but as they gain the begins to push back and their acceleration decreases (they are still gaining speed each second, they are just not gaining as much speed each second as they were at the start). Finally the drag force becomes as large as the weight and their acceleration becomes zero, and they have reached . In the next chapter we will work on understanding how to use and to predict the of objects, including our skydiver, and how graphing the position, velocity, and acceleration can help us do so.

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