Acceleration, velocity, and Position

The connections between position, velocity, and acceleration formed one of the important themes of differential calculus. We will find that these relationships also form an important application of the definite integral, especially in cases in which one of the quantities varies with time.
To discuss these concepts, we will use the notation:

$x(t) = ~~{\rm position~ at~ time~} t \] \[ v(t) = ~~{\rm velocity~ at~ time~} t \] \[ a(t) = ~~{\rm acceleration~ at~ time~} t$

Relating velocity to acceleration
Remembering that the acceleration is defined by the derivative
$a(t) = \frac{dv}{dt}$
we can apply the Fundamental Theorem of Calculus to write this relationship in the form
$\int_{T_1}^{T_2}~~a(t) ~dt= ~v(T_2)~-~v(T_1)$
If we pick call the initial time and the final time , then this integral has the form
$\int_{0}^{T}~~a(t) ~dt= ~v(T)~-~v(0)$

Relating Position to Velocity
The velocity is defined by the derivative
$v(t) = \frac{dx}{dt}$
By the Fundamental Theorem of Calculus,
$\int_{T_1}^{T_2}~~v(t) ~dt= ~x(T_2)~-~x(T_1)$
We call this difference in the final and the starting positions, , the displacement.
As before, we can chose to start the clock so that the process takes place between and , so that the integral becomes:
$\int_{0}^{T}~~v(t) ~dt= ~x(T)~-~x(0)$

Example 1: The case of constant acceleration
When the acceleration is constant, , we can use the general results above to compute an expression for the velocity and the position at a given time. Here is how we would do this. First to find the velocity at time t, we would integrate the acceleration:
$\int_{0}^{T}~~a(t) ~dt= ~v(T)~-~v(0) \] \[ a~ t |^T_0= ~v(T)~-~v(0) \] \[ a~ T= ~v(T)~-~v(0)$
This implies that

$v(T) = v(0) + a T$

Now, to find the position, we integrate this velocity, as follows:
$\int_{0}^{T}~~v(t) ~dt= ~x(T)~-~x(0) \] \[ \int_{0}^{T}~~(v(0) + a t) ~dt= ~x(T)~-~x(0) \] \[ (v(0) t + \frac12 ~a t^2) |^T_0 = ~ x(T)~-~x(0) \] \[ v(0) T + \frac12 ~a T^2 = ~ x(T)~-~x(0)$
Rearranging algebraically, leads to the result

$x(T) = x(0) + v(0) T + \frac12 ~a T^2$

A simple application (constant acceleration)

We can apply the general result above to the following simple problem:

The speed of a car increases from rest up to 100 km/hr in 30 seconds. Find the distance that the car has travelled during this time.

To solve this problem, we first observe that the initial velocity is , and the final velocity is . Also note that . Care must be taken to use consistent units for time and distance here.
Since the acceleration is constant we might observe that, for this case,
$a(t) = \frac{dv}{dt} = \frac{\Delta v}{\Delta t} = \frac{100}{30} \frac{1000}{3600} = 0.92 ~{\rm m/s}^2$
(where the fraction 1000/3600 has been used to convert the kilometres to metres and the hours to seconds). From the previous results we can now express the velocity at some time t as
$v(t) = v(0) + a t = 0 + 0.92 t = 0.92 t$
We now find the displacement
$D(T) =\int_0^{30} v(t) dt = \int_0^{30} 0.92 t dt = 0.92 \int_0^{30} t dt = 0.92 \frac{t^2}{2}|_0^{30} = 0.92 \frac{900}{2} =414 ~ m$
Thus the car has travelled 414 metres during the first 30 seconds when it is accelerating

For your consideration:

(1) Suppose that at the car is travelling at velocity and that the brakes is applied so that it has stopped after 30 seconds. Find the deceleration and the distance that the car has travelled during that time.

(2) Suppose that you are driving your car at 100 km/hr when a pedestrian dashes across the street. It takes you 0.5 second to react (during which the car continues to move forward at the same speed). You then slam on the brakes and decelerate at the rate . How far will the car move before coming to a complete stop ?

Example 2: cases in which the acceleration is not constant
When a skydiver jumps from an airplane, once the parachute is open, the speed of falling "levels off" to some constant, safe speed, called the terminal velocity. We have seen last term that the velocity of the skydiver can be described by the differential equation
$\frac{dv}{dt} = g - k v$
where is acceleration due to gravity, and is the slowing down due to the friction between the parachute and the air. You may or may not remember the details, but the conclusion we had found by studying this differential equation was that that
$v(t) = \frac{g}{k}(1- e^{- kt})$

For your consideration:

(1) Take the derivative of this expression. Show that the expression does indeed satisfy the differential equation for v.
(2) Find out what this statement about the velocity implies about the acceleration, that is, find a differential equation that the acceleration satisfies.
(3) What is the solution of the differential equation for the acceleration? (Hint: remember the differential equation for exponential decay)
(4) Using your answer in (3), you should be able to now show that the expression for the velocity given above is the correct one. Integrate to show this.

At present, even if the details that lead to this expression for the velocity have been forgotten, we can use this example as an illustration of a calculation of displacement in the non-constant acceleration case. We will simply note the velocity is approaching a constant level for large , namely, $v(t) \approx g/k$ . We can still find the displacement using our integration procedure, as follows:
${\rm Displacement ~} = D(t)= x(T) - x()) =\int_{0}^{T}~~v(t) ~dt \] \[ D(t) = \int_{0}^{T}~~\frac{g}{k}(1- e^{- kt}) ~dt = \frac{g}{k}\int_{0}^{T}~(1- e^{- kt}) ~dt \] \[ D(t) = \frac{g}{k} [ t + \frac{e^{-kt}}{k}]|^T_0 \] \[ D(t) = \frac{g}{k} [ T + \frac{e^{-kT}-1}{k}]$
Observe that for large T, the exponential term in the above expression is negligible, so that
$D(t) = \frac{g}{k} [ T + \frac{-1}{k}] = \frac{g}{k}T - \frac{g}{k^2}$