# Velocity and Acceleration questions

1. Sep 14, 2015

### Dillion

• Missing template. Originally posted in different forum.
x(t)= -4m-(9m/s)t+(2m/s)t^2

I'm lost on velocity and acceleration. Below are questions on a worksheet and i answered positive and negative. Can someone correct me on them please? Thank you!

Is the initial acceleration in the positive x or negative x direction?
positive?
Is the initial velocity in the positive x or negative x direction?
negative?
Is the initial position in the positive x or negative x direction from the origin?
negative?
After 3 seconds is the acceleration in the positive x or negative x direction?
positive?
After 3 seconds, is the velocity in the positive x or negative x direction?
negative?
After 3 seconds is the position in the positive x or negative x direction from the origin?
negative?

2. Sep 14, 2015

### jbriggs444

If t is in seconds then that formula is dimensionally incorrect. What would the resulting units be on the acceleration term?

3. Sep 14, 2015

### Dillion

x(t)= -4m-(9m/s)t+(2m/s^2)t^2

I forgot to square the acceleration...does that help?

4. Sep 14, 2015

### jbriggs444

But... this should have been posted in the homework section using the supplied template. And you need to show some work. You have a function for position as a function of time. Start with the third question. Can you answer that?

5. Sep 14, 2015

### Dillion

I want to say negative because of the -4m...but I have a feeling that is wrong

6. Sep 14, 2015

### jbriggs444

That would be correct.

If you evaluate the expression $-4{meters} -9\frac{meters}{sec}t + 2\frac{meters}{{sec}^2}t^2$ for $t$ = 0 sec then the result would be -4 meters.

Velocity is the rate of change of position with respect to time. If t is near zero and is increasing, what can we say about x? Is x increasing or decreasing? Is there a term in the equation that can tell you that easily?

If you have to, try evaluating the expression for t = 0.01 and see what the result is.

7. Sep 14, 2015

### Dillion

so when answering all the "initial" questions, I just have to look at the sign in the equation.

When it asks for "After 3 seconds", I would still have to look at the signs for acceleration and velocity but not position, correct? For position, I could just plug in 3 seconds into t and solve.

8. Sep 14, 2015

### jbriggs444

By this, I assume that you mean "look at the sign of the corresponding term". That works if t is zero.
Yes, for position, you can just plug in 3 seconds and evaluate the formula.

But for velocity... You can read off $-9 \frac{meters}{sec}$ for the velocity at t=0. But by t=3, acceleration has occurred. The new velocity is not the same as the initial velocity.