#### Math problems and solution > Derivatives > Basic examples > Solved exercises > Intermediate level > Solution 6

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### 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 :

• 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) = 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 If y = 0 we have In that case, the value of x, x1 and x2, can be calculated as follow:

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.