# Acceleration & max displacement from velocity equation

1. Mar 16, 2017

### ldesai149

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

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

3. The attempt at a solution

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!

2. Mar 16, 2017

### TSny

For (b) and (c) you are using "constant acceleration" equations. Is the acceleration of this object constant?

3. Mar 16, 2017

### ldesai149

The question doesn't say either way... I get the feeling I'm missing something but I don't know what.

4. Mar 16, 2017

### TSny

Use the given expression for the velocity as a function of time to decide if the acceleration is constant.

5. Mar 16, 2017

### ldesai149

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. Mar 16, 2017

### TSny

Right.
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. Mar 16, 2017

### ldesai149

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. Mar 16, 2017

### TSny

OK. But what is the value of C?
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. Mar 16, 2017

### ldesai149

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. Mar 16, 2017

### TSny

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.

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