Velocity and acceleration derivatives

[corn_she]

I have another question..

The velocity and acceleration are the first and second derivatives of distance (x).

The distance covered by a vehicle is given by x=4t^3 - 8t^2 + t - 7

Calculate the velocity at time= 2 seconds.

Calculate the acceleration at time = 1.3 seconds.

Find the turning point of the velocity and explain what is happening to the car at that point.

My working

x = 4t^3 - 8t^2 + t - 7
v=dx/dt = 12t^2 - 16t + 1 - 7 Velocity
dv/dt = a = d^2x/dt^2 = 24t - 16 + 1 -7 Acceleration

12.2^2 - 16.2^2 + 1 - 7 = -22 metres per second velocity
24.1.3 - 16 +1 - 7 = 9.2 metres per second acceleration

Is that correct?

How do i find the turning point?

 

[Deleted member 4993]

I have another question..

The velocity and acceleration are the first and second derivatives of distance (x).

The distance covered by a vehicle is given by x=4t^3 - 8t^2 + t - 7

Calculate the velocity at time= 2 seconds.

Calculate the acceleration at time = 1.3 seconds.

Find the turning point of the velocity and explain what is happening to the car at that point.

My working

x = 4t^3 - 8t^2 + t - 7

v=dx/dt = 12t^2 - 16t + 1 - 7 Velocity ..........Incorrect ............ dx/dt = 12t^2 - 16t + 1

dv/dt = a = d^2x/dt^2 = 24t - 16 + 1 -7 Acceleration..........Incorrect ............ d^2x/dt^2 = 24t - 16

12.2^2 - 16.2^2 + 1 - 7 = -22 metres per second velocity
24.1.3 - 16 +1 - 7 = 9.2 metres per second acceleration

Is that correct? ..........Incorrect

How do i find the turning point?

.

 

[ksdhart]

Okay, so you have a formula for finding the velocity at any given moment. That's good, except for one thing. When taking the derivative, you have

Velocity: v=dx/dt = 12t^2 - 16t + 1 - 7

But that's not quite right. Think about what you're really doing when taking the derivative. x = 4t^3 - 8t^2 + t - 7. So then the derivative of x, by the addition rule is:

$\displaystyle \frac{d}{dt}\left(4t^3\right)-\frac{d}{dt}\left(\:8t^2\right)+\frac{d}{dt}\left(t\right)-\frac{d}{dt}\left(7\right)$

You calculated the derivative of 7 as being 7. Can you see why that's not right?

As for your next question "Find the turning point of the velocity and explain what is happening to the car at that point." A good start is to define turning point. I think the act of typing out what it means will help you figure out what to do. And if not, we'll give you a hint.

 

[Steven G]

I have another question..

The velocity and acceleration are the first and second derivatives of distance (x).

The distance covered by a vehicle is given by x=4t^3 - 8t^2 + t - 7

Calculate the velocity at time= 2 seconds.

Calculate the acceleration at time = 1.3 seconds.

Find the turning point of the velocity and explain what is happening to the car at that point.

My working

x = 4t^3 - 8t^2 + t - 7
v=dx/dt = 12t^2 - 16t + 1 - 7 Velocity
dv/dt = a = d^2x/dt^2 = 24t - 16 + 1 -7 Acceleration

12.2^2 - 16.2^2 + 1 - 7 = -22 metres per second velocity
24.1.3 - 16 +1 - 7 = 9.2 metres per second acceleration

Is that correct?

How do i find the turning point?

You really should have replaced 1-7 with -6.
The derivative function gives the slope of the tangent line of a function. If the function is a constant, then at any point on the line the slope of the tangent line is 0. That is the derivative of 7 is 0, not 7.
The turning point of a function f(x)is at an x-value where f'(x) changes signs. A necessary condition for a turning is that f'(x)= 0. For example if f'(x)= x^2, then the f'(x)= 0 when x=0. If x is a 'little' than 0 or x is a 'little' more than 0 f'(x) does not have a sign change-so no turning point for f(x). If however the f'(x)= x^3, then there is a sign change around 0 and the function does have a turning point for f(x) at x=0

 

[corn_she]

v= 12t^2 - 16t + 1
a = 24t - 16

12.2^2 - 16.2^2 + 1 = 17 metres per second velocity
24.1.3 - 16 = 15.2 metres per second acceleration

I know that the turning point is where the line crosses the x axis.

What do i do next?

 

[corn_she]

is it dy/dx = 12t^2 - 16 t + 1
= 24t -16
24/16 = 1.5 seconds

 

[stapel]

I know that the turning point is where the line crosses the x axis.

Really? When you're working with "x(t)" and differentiating with respect to t, what does it mean to "cross the x-axis"? Will you, in fact, ever "cross the x-axis"? Or, graphically, are you working along the x-axis, with the values of t telling you where you are?

When you're dealing, conceptually, with position and velocity, what does the "turning point" of the position function tell you, with respect to the velocity? Think of a (generic) position graph. Think of its ups and down and, specifically, its "bumps" (its max and min points). Where are the turning points of this position graph? How do these relate to the slope of the position graph and thus to the velocity function?

How are velocity graphs, their turning points, and acceleration related?

How can this be applied to your exercise?

 

[Deleted member 4993]

is it dy/dx = 12t^2 - 16 t + 1

= 24t -16 ............... This is equal to what?

24/16 = 1.5 seconds ....... Where did this come from?

What are you doing??!!

Where is 'y' and 'dy/dx' coming from? You are dealing with a function of 't' → x(t)