Acceleration & max displacement from velocity equation

In summary: What is that location?The object is at the origin, so x(0) = 0. Can you use this to determine the value of C?Yes, that makes sense. So C = 0 and x(t) = -1/3t3 + t2 for t=3.0s is -18.0m. And the maximum positive displacement is still at t=2.0s with a value of 1.33m.In summary, the velocity of an object moving along the x-axis is given by Vx = (2.0t-t2)m/s. Initially (at t=0), the object was at the origin. To determine the object's acceleration at t=
  • #1
ldesai149
8
0

Homework Statement



The velocity of an object moving along the x-axis is given by Vx = (2.0t-t2)m/s. Initially (at t=0), the object was at the origin.

a) Determine the object's acceleration at t=3.0s.
b) Find the object's position at t=3.0s
c) Calculate the object's maximum positive displacement from the origin.

Homework Equations



a = dv/dt
xf = xi + vxit + (1/2)axt2
vxf2 = vxi2 + 2ax

The Attempt at a Solution


[/B]
a) To get an equation for acceleration,

ax = dv/dt = d/dt (2.0-2t)
= 2.0 - 2(3.0s)
= -4.0 m/s2

b) xf = xi + vxit + 1/2axt2
= (0 m) + (0 m/s)(3.0 s) + 1/2 (-4.0 m/s2)(3.0s)2
= -18.0 m

c) delta x = [vxf2 - vxi2] / [2ax]
= [2.0 (3.0s) - (3.0s)2]2 - [0 m/s] / [2 (-4.0 m/s2)]
= -1.1 m

First of all, I'm not sure if I've done the derivation for the accleration equation correctly. Also, question c) asks for maximum POSITIVE displacement but I keep getting a negative value. I'm not sure where I'm going wrong!
 
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  • #2
For (b) and (c) you are using "constant acceleration" equations. Is the acceleration of this object constant?
 
  • #3
The question doesn't say either way... I get the feeling I'm missing something but I don't know what.
 
  • #4
Use the given expression for the velocity as a function of time to decide if the acceleration is constant.
 
  • #5
Ok, so if I plug in different values for t in acceleration equation, the values are not constant... The only solution I can think of is to calculate x separately for each second? Also, am I correct in assuming that my answer for b) is still accurate since I calculated acceleration for the same time (t=3.0s)?

Thank you for all your help! (I'm a healthcare student who's forgotten all her high school physics...)
 
  • #6
ldesai149 said:
Ok, so if I plug in different values for t in acceleration equation, the values are not constant...
Right.
The only solution I can think of is to calculate x separately for each second? Also, am I correct in assuming that my answer for b) is still accurate since I calculated acceleration for the same time (t=3.0s)?
Can you derive the position vs time function x(t) from the velocity vs time v(t)? What calculus operation should you use to do this?
 
  • #7
Alright, so I integrated v(t) to get x(t) = -1/3t3 + t2 + C.
Input values of t=1,2,3,4 and found that at t=2.0s, the positive displacement is 1.33m, at t=3, it is 0m and at t=4, it is -5.3m.
So, the maximum positive displacement is 1.33m at t=2.0s.

Have I finally got it right?
 
  • #8
ldesai149 said:
Alright, so I integrated v(t) to get x(t) = -1/3t3 + t2 + C.
OK. But what is the value of C?
Input values of t=1,2,3,4 and found that at t=2.0s, the positive displacement is 1.33m, at t=3, it is 0m and at t=4, it is -5.3m.
So, the maximum positive displacement is 1.33m at t=2.0s.
Yes, at t = 3.0 s, the object is at x = 0.

For part (c), how do you know that the maximum positive value of x will occur for an integer value of t?
 
  • #9
When I solve for C:

From b), I have x(t) for t=3.0s is -18.0m

x(3) = -18.0 = (3)2 - (1/3)(3)3 + C
-18.0 = C

But that doesn't seem right.
 
  • #10
In your first post, you didn't get the correct answer for (b) because you assumed that the acceleration is constant.

You can determine the value of C by using the information given in the problem statement about the location of the object at t = 0.
 

Related to Acceleration & max displacement from velocity equation

What is the equation for acceleration?

The equation for acceleration is a = (vf - vi) / t, where a is acceleration, vf is final velocity, vi is initial velocity, and t is time.

How do you calculate maximum displacement from velocity?

To calculate maximum displacement from velocity, you can use the equation x = (vf^2 - vi^2) / 2a, where x is maximum displacement, vf is final velocity, vi is initial velocity, and a is acceleration.

What is the difference between acceleration and velocity?

Acceleration is the rate of change of velocity over time, while velocity is the rate of change of position over time. In other words, acceleration measures how quickly an object's velocity changes, while velocity measures how quickly an object's position changes.

What units are used to measure acceleration?

Acceleration is typically measured in meters per second squared (m/s^2) in the metric system, or feet per second squared (ft/s^2) in the imperial system.

Can acceleration be negative?

Yes, acceleration can be negative. This means that an object is slowing down, or its velocity is decreasing over time. Negative acceleration is also known as deceleration.

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