velocity is minimum,what is dv/dt?
Answers
Answer:
Application: Kinematics
Aims: By the end of this chapter, you will be able to
(i) interpret kinematic questions and set up the appropriate mathematics,
(ii) apply differentiation to solve kinematic questions.
Basically, in this topic we will study various relationships between displacement and velocity, displacement and acceleration and acceleration and velocity.
Velocity (v) measures the rate of change of displacement. In another word, v = ds / dt where s is the displacement and t is time.
Interpreting Signs in Kinematic Questions.
Here we will suppose there is an object P that moves in a straight line with displacement s(t) with respect to the origin O at (0,0). Its velocity function and acceleration functions are v(t) and a(t) respectively.
s = 0
s > 0
s < 0
P is at origin O
P is to the right of O
P is to the left of O
v = 0
v > 0
v < 0
P is instantaneously at rest (P could be at a stationary point, at a point of reversing direction, or reaches the max displacement.)
P is to moving to the right (forward)
P is to moving to the left (backward)
a = 0
a > 0
a < 0
velocity may be maximum or minimum
velocity is increasing over time
velocity is decreasing over time
Making sense of questions:
* initial condition or at the start of the experiment
* object is at the origin
* object is at a stationary position, reverses direction, or at maximum displacement (height)
* object moving at constant velocity or reaches max/min velocity
t = 0
s = 0
v = 0
a = 0
Example 1: If the displacement of an object can be modelled as
s = ut + kt2 where u and k are some constants then
v = ds/dt = u + 2kt
Acceleration (a) measures the rate of change of velocity. In another word, a = dv / dt.
Also a = dv / dt
a = d2s / dt2 this is the second derivative of displacement.
Acceleration can also be expressed as a = (dv / ds)(ds / dt) and rearranging it yields
a = (ds / dt)(dv / ds)
a = v (dv / ds)
Using the previous example 1, s = ut + kt2 , the acceleration a = 2k in this particular case, acceleration is a constant. However, acceleration can also be a function of time as in the following example.
Example 2: The velocity (V) of a certain particle is V = 80 t - (1/4)t2 where t is time measured in seconds and V in cm/s. Find the acceleration of this particle at t =1 minute.
Solution:
a = dV/dt = 80 - (1/2)t. In this case, acceleration is a function of time, t.
a(60) = 80 - (1/2)(60) ------------ we need to change the 1 minute into 60 seconds as t is measured in seconds.
a(60) = 50 cm/s2
