Displacement, velocity, acceleration

In summary: Sorry about that. In summary, the particle's displacement during the first 3 seconds is 4 ft, its average velocity during the first 3 seconds is 1 ft/s, its instantaneous velocity when t = 3 is 19 ft/s, its acceleration when t = 3 is 18 ft/s^2, and it changes direction at t = sqrt(8/3)s or 1.633333s.
  • #1
Jacobpm64
239
0
A particle moves along a line so that its position at any time t>0 is given by the funciton s(t) = t^3 - 8t + 1, where s is measured in feet and t is measured in seconds.
(a) Find the displacement during the first 3 seconds.
(b) Find the average velocity during the first 3 seconds.
(c) Find the instantaneous velocity when t = 3.
(d) Find the acceleration of the particle when t = 3.
(e) At what value or values of t does the particle change direction?

Ok, this is pretty much just to check my answers because I'm paranoid.. Here are the answers i came out with..

(a) 4 ft
(b) 1 ft/s
(c) 19 ft/s
(d) 18 ft/s^2
(e) sqrt(8/3)s or 1.633333s

Is this correct?

Thanks
 
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  • #2
Jacobpm64 said:
(b) 1 ft/s
Check this one again. Other than that the answers seem fine.
 
  • #3
LeonhardEuler said:
Check this one again. Other than that the answers seem fine.

Yea except for that one I think they're all right.
 
  • #4
let's see.. for that one i did..

s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s

not sure on this? can you point me in the right direction?
 
  • #5
Jacobpm64 said:
let's see.. for that one i did..
s(t) = t^3 - 8t + 1
s(0) = 0^3 - 8(0) + 1 = 1
s(3) = 3^3 - 8(3) + 1 = 4
so it would be
average velocity = change in displacement/change in time
average velocity = (4 - 1) / (3 - 0) = 3/3 = 1 ft/s
not sure on this? can you point me in the right direction?

Ohh yea I guess you're right I just wasn't thinking right and thought that initial displacement was zero, you're doing it right.
 
  • #6
Uhh, yeah never mind-oops!
 

What is displacement?

Displacement is the change in position of an object from its initial point to its final point. It is a vector quantity, meaning it has both magnitude and direction. It can be calculated by subtracting the initial position from the final position.

What is velocity?

Velocity is the rate of change of an object's displacement over time. It is a vector quantity and is calculated by dividing the change in displacement by the change in time. It is often represented by the symbol "v" and its units are meters per second (m/s).

What is acceleration?

Acceleration is the rate of change of an object's velocity over time. It is a vector quantity and is calculated by dividing the change in velocity by the change in time. It is often represented by the symbol "a" and its units are meters per second squared (m/s^2).

How are displacement, velocity, and acceleration related?

Displacement, velocity, and acceleration are all related through the equations of motion. Velocity is the derivative of displacement with respect to time, and acceleration is the derivative of velocity with respect to time. This means that acceleration can be calculated by taking the second derivative of displacement with respect to time.

What are some real-world examples of displacement, velocity, and acceleration?

Some real-world examples of displacement, velocity, and acceleration include a car driving down the highway (displacement is the distance traveled, velocity is the speed at which it is traveling, and acceleration is how quickly its speed is changing), a ball thrown into the air (displacement is the height it reaches, velocity is its speed at any given point, and acceleration is the force of gravity pulling it back down), and a rollercoaster ride (displacement is the length of the track, velocity is the speed of the coaster at any point, and acceleration is the force that pushes the coaster along the track).

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