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) = 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 |
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.