Velocity and Acceleration questions

[Dillion]

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?

 

[jbriggs444]

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

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

 

[Dillion]

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

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

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

 

[jbriggs444]

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

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

Yes, that addresses that problem.

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?

Is the initial position in the positive x or negative x direction from the origin?
negative?

 

[Dillion]

Yes, that addresses that problem.

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?

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

 

[jbriggs444]

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

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.

 

[Dillion]

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.

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.

 

[jbriggs444]

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

By this, I assume that you mean "look at the sign of the corresponding term". That works if t is zero.

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.

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.