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The DErivatives

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THE

DERIVATIVES
OBJECTIVE:
APPLY THE DERIVATIVE CONCEPT
IN THE STUDY OF MOTION
Kinematics is the study of motion and is closely related to
calculus. Physical quantities describing motion can be
related to one another by derivatives.
1. DISPLACEMENT IS THE SHORTEST DISTANCE
BETWEEN TWO POSITIONS AND HAS A DIRECTION.

Example:
• The park is 5 kilometers north of here
• x(t) =5(t), where x is displacement from a point P
and t is time in seconds.
2. VELOCITY REFERS TO THE SPEED AND
DIRECTION OF AN OBJECT.

Example:
• Object moving 5 m/s backwards
• v(t)= , where v is an object’s velocity and t is time in
seconds
3. ACCELERATION IS THE RATE OF CHANGE OF VELOCITY PER UNIT TIME.
IMAGINE INCREASING YOUR SPEED WHILE DRIVING. ACCELERATION IS
HOW QUICKLY YOUR SPEED CHANGES EVERY SECOND.

Example:
• Increasing speed from 10 m/s to 25 m/s in 5 s results in:

• a(t)=-t, where a is an object’s acceleration and t is time in seconds


GIVEN X(T) AS DISPLACEMENT, V(T) AS VELOCITY AND
A(T) AS ACCELERATION, WE CAN RELATE THE
FUNCTIONS THROUGH DERIVATIVES.

a ( t ) = 1 l( t ) = x 11
( t )

• Equivalently, using Leibniz notation:



ch, d 2 x
.a(t) = -= ­
dt dt 2
X

The maximum of the


displacement function,
x(t), occurs along the
flat blue line where the
t
8 12
rate of change is zero.
EXAMPLE 1:
If a particle is moving in space with a velocity function, v(t)= -2t-8
where t is seconds and velocity is measured in meters per
second.
(a) At what time(s), if any, is the particle at rest?
(b) What is the acceleration of the particle at t=3 seconds?
SOLUTION:
(a)At what time(s), if any, is the • Since negative time is
particle at rest? impossible, the only time at
which the particle is at rest is 4
• (a) If the particle is at rest, v(t)=0 seconds.
(velocity is zero at rest)
V
Solving for t when v(t) = 0:
SOLUTION:
(B)WHAT IS THE ACCELERATION OF THE PARTICLE AT
T=3 SECONDS?

• First find the function for • Thus, the acceleration at t = 3 s


acceleration by taking the is 4 m/s2
derivative of velocity.

Substitute t = 3 s in the
acceleration function: a(3) = 2(3) -
2 = 4 m/s2
J,.
Example 2 :
A soccer ball is kicked into the air so that the path of its flight
can be modeled by the function, where t is in seconds and x
is meters above ground:

𝑥 ( 𝑡 ) =− 4.9 𝑡 2 +9.8 𝑡 +5

(a)At what time will the ball land?


(b)How many meters above ground was the ball kicked?
(c)What is the maximum height the ball will reach and at what time will
this occur?
(d)What is the acceleration (with direction) of the ball at t=3 s?
SOLUTION:
(A) AT WHAT TIME WILL THE BALL LAND?

• Since x(t) models height above


ground, x(t)=0 when the ball hits the
ground

• However, t is greater than 0, (since


time cannot be negative). Thus, the
ball hits the ground 2.421 seconds
after being launched.

-2.421
(B)HOW MANY METERS ABOVE GROUND WAS THE BALL
KICKED?

• The initial height above ground occurs when t = 0. Substitute t = 0


into x(t):

• Thus, the ball is thrown from 5 meters above ground.


(C)WHAT IS THE MAXIMUM HEIGHT THE BALL WILL
REACH AND AT WHAT TIME WILL THIS OCCUR?

• Maximum height occurs


when the first derivative
equals zero.
(D)WHAT IS THE ACCELERATION
(WITH DIRECTION) OF THE BALL
AT T=3 S?

• Acceleration is equal to the


second derivative of
t
displacement.
THANK YOU!!!

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