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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.

Derivatives : solution exercise 6 (Intermediate math level)


A particle is moving along a straight line. Its position x with respect to time is given by the equation x(t) = 2t3 - 12t2 + 20t + 3 where x(t) represents the distance (in meters) traveled by the particle at time t. You are asked to :


This exercise is pretty much like the exercise 5.

x(t) = 2t3 - 12t2 + 20t + 3 where x(t) is in meters and t in seconds.

x '(t) = v(t) = 6t2 - 24t + 20 where v(t) is in m/sec.


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 t = 2sec , the particle will be at a distance of 11 m from the point of departure.

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 4 m/s.


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) = 6t2 - 24t + 20

2 m/s = 6t2 - 24t + 20

6t2 - 24t + 18 = 0

Reminder:

Given the second degree equation Second degree equations are equations of a parabola. The derivative of the velocity v(t) with respect to time yields the instantaneous accelaration a(t) of a particle
If y = 0 we have the first derivative of the position (or distance traveled) with repsect to time, x'(t), yields the instantaneous velocity v(t) of a particle
In that case, the value of x, x1 and x2, can be calculated as follow: Solved physics problems online for free, Uniformly Accelerated Motion, non-uniform motion, first derivative of position, derivative position, first derivative of velocity, equation of the traveled distance as a function of time

mechanics and straight line motion, physics, Velocity is the first derivative of the position. Acceleration is a rate at which a velocity is changing. Acceleration is a rate of change

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.


Derivatives : Intermediate math level exercises