Acceleration & max displacement from velocity equation

[ldesai149]

Homework Statement

The velocity of an object moving along the x-axis is given by V_x = (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
x_f = x_i + v_xit + (1/2)a_xt²
v_xf² = v_xi² + 2a_x

The Attempt at a Solution

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

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

b) x_f = x_i + v_xit + 1/2a_xt²
= (0 m) + (0 m/s)(3.0 s) + 1/2 (-4.0 m/s2)(3.0s)²
= -18.0 m

c) delta x = [v_xf² - v_xi²] / [2a_x]
= [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!

 

[TSny]

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

 

[ldesai149]

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

 

[TSny]

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

 

[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...)

 

[TSny]

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?

 

[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?

 

[TSny]

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?

 

[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.

 

[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.