**
Online physics exercises with solution : mechanics and straight line motion. The velocity is the first derivative
of the position. The accelaration is a rate at which a velocity is changing. You will find here solved physics problems that use the derivatives and mathematics in physics.
**

A particle is moving along a straight line. Its position *x*
with respect to time is given by the equation *x(t) = 2t ^{3} - 12t^{2} + 20t + 3*
where

a) Find the position of the particle at time

*t = 2 sec*and to figure out its instantaneous velocity at that instant.b) Find out the position of the particle when its instantaneous speed is

*v = 2 m/sec*.

This exercise is pretty much like the exercise 5.

*x(t) = 2t ^{3} - 12t^{2} + 20t + 3* where

*x '(t) = v(t) = 6t ^{2} - 24t + 20* where

**a) Find the position of the particle at time t = 2 sec and figure out its instantaneous velocity at that instant**

The position after 2 seconds will be *x(2) = 2.(2) ^{3} - 12.(2)^{2} + 20 . 2 + 3 = 11 m*. Thus at time

* v(2) = 6.(2) ^{2} - 24.(2) + 20 = -4 m/s*. Notice that the particle has a negative speed. That means that the particle is going backward with an absolute velocity of

**b) Find out the position of the particle when its instantaneous speed is v = 2 m/sec**

What is the position of the particle when its velocity is *v = 2 m/sec *?

*v(t) = 6t ^{2} - 24t + 20*

*2 m/s = 6t ^{2} - 24t + 20*

*6t ^{2} - 24t + 18 = 0*

Reminder:

Given the second degree equation x, can be calculated as follow: _{2} |

That means that the instantaneous speed is *2 m/s* at both times * t = 3 sec* and *t = 1 sec*. But what will be the position around the point of departure at both times ? I mean the particle will be farther after the point of departure or back behind the point of departure ?

At *t = 1 sec*

*x(1) = 2 (1) ^{3} - 12 (1)^{2} + 20 (1) + 3 = 13 meters* farther after the point of departure.

At *t = 3 sec*

*x(3) = 2 (3) ^{3} - 12 (3)^{2} + 20 (3) + 3 = 9 meters* farther after the point of departure.