**
Uniformly Accelerated Motion : free online physics exercise with complete detailed solution.
The derivative of the position as a function of time gives the velocity of a particle. The derivative of the
distance traveled by a particle as a function of time yields the instantaneous speed (or instantaneous velocity).
**

### Physics, derivatives and Uniformly Accelerated Motion :

solution exercise 5 (Intermediate math level)

A particle is moving along a straight line *X* according to the following equation

*x(t) = 2t*^{2} + 8t + 9, where *x(t)* is the distance (in meters) traveled by the particle
at time *t*, calculated from the point of departure. You are asked to find the position of the particle and its velocity as well :

at time *t = 0 sec*

at time *t = 1 sec*

Notice that you have here the equation of a moving object that is speeding up along a straight line.
This is a non-uniform motion because the instantaneous velocity is changing all the time, as a result
of which the object is said to have an acceleration. In physics, it is called *Uniformly Accelerated
Motion*. An example of a *Uniformly Accelerated Motion* would be a car accelerating on a straight highway
(notice that the accelaration is 4 m/s^{2} in our exercise because the factor of t^{2} is 2. And we know that, in physics, that factor times two yields the value of the accelaration).
To find out the traveled distance by that particle (let's say it is a car) or that car, at various instants,
it is quite easy, you just need to replace *t* in the equation *x(t)* by the value of time that you are considering.
This way, at time *t = 0 sec* (which means when the driver is starting the motor of its car and is stepping on the accelarator),
we observe that the car is already at a distance

*x(0) = 2 (0)*^{2} + 8 (0) + 9 = 9 meters from the point of departure.
And at time *t = 1 sec*, the car is at a distance *x(1) = 2 (1)*^{2} + 8 (1) + 9 = 19 meters from the point of departure.
You know now that *x(t) = 2t*^{2} + 8t + 9 is what we call the equation of the position (or traveled distance) as a function of time. And the derivative of that equation yields a new function *v(t)* which is the
instantaneous velocity (or instantaneous speed) as a function of time:

*x '(t) = v(t) = 4t + 8*

To figure out the instantaneous velocity of the car at time *t = 0 sec*, you just need to replace *t* by *0* in the equation *v(t)*.
Thus at *t = 0 sec*, *v(0) = 4 (0) + 8 = 8 m/s*. You can see that when the motor of the vehicle is started, the car already has a speed greater than *0*. This means that our car has let itself go down the street with its motor turned off and without applying the breaks. The chronometer has been started later, immediately after the driver had started the motor and stepped on the accelarator.

And at *t = 1 sec *, *v(1) = 4 (1) + 8 = 12 m/s*. Thus in one second the car has accelarated from 8 m/s to 12 m/s (which gives a mean accelaration of *[12m/s - 8m/s]/1sec = 4 m/s*^{2} which is the same value that the one calculated previously hereabove).