Finding the slope of this function(the slope is also avg. velcoity I think)

A car is stopped at a traffic light. It then travels along a straight road so that its distance from the light is given by http://session.masteringphysics.com/...t%5E2+-+ct%5E3, where http://session.masteringphysics.com/render?var=b = 2.40 http://session.masteringphysics.com/...nits=m%2Fs%5E2 and http://session.masteringphysics.com/render?var=c = 0.100 http://session.masteringphysics.com/...nits=m%2Fs%5E3.

1. Calculate the average velocity of the car for the time interval http://session.masteringphysics.com/render?tex=t+= 0 to http://session.masteringphysics.com/render?tex=t+= 10.0 http://session.masteringphysics.com/...=%7B%5Crm+s%7D.

2. How long after starting from rest is the car again at rest?

I know this is borderline physics problem, but I need to find the slope of this function, which will give me the average velocity, but I do not really know how.

When the time is 0, I received 2.3 meters.

When the time is 10, I received 340 meters.

However, when I did 337.7 meters divided by 10 seconds, I didn't get the right answer. Help is needed! thanks!

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**CrusaderKing1**

1. how did you get x(0) = 2.3 ?

2. find when $\displaystyle \frac{dx}{dt} = 0$

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**skeeter** 1. how did you get x(0) = 2.3 ?

2. find when $\displaystyle \frac{dx}{dt} = 0$

So use the derivative of the function(which is 2bt+3ct^2) and plug in 5.00 s(a point besides 0) and 10.0 so I get a slope of 0.3 by (5,16.5) and (10,18)?

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**CrusaderKing1** So use the derivative of the function(which is bt+3ct^2) and plug in 5.00 s(a point besides 0) and 10.0 so I get a slope of 0.3 by (5,16.5) and (10,18)?

huh?

$\displaystyle v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

the derivative of the position function is **instantaneous velocity** ... you do not need a derivative to find **average velocity**.

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**skeeter** huh?

$\displaystyle v_{avg} = \frac{x(t_2) - x(t_1)}{t_2 - t_1}$

the derivative of the position function is **instantaneous velocity** ... you do not need a derivative to find **average velocity**.

t2 = 10.0

t1= 0.00

x(t2)= 140

140/10 =14 ?

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**CrusaderKing1** t2 = 10.0

t1= 0.00

x(t2)= 140

140/10 =14 ?

14 m/s ... that's it, Zeke.

Re: Finding the slope of this function(the slope is also avg. velcoity I think)

Quote:

Originally Posted by

**skeeter** 14 m/s ... that's it, Zeke.

Thanks a billion man! Ya, I had 3 questions on instantaneous velocity and I got them all right fairly fast, but average velocity was difficult for me. If I have a question on the last problem, ill ask again, but im going to try to figure it out first.