Questions on kinematics (distance/velocity/acceleration)

[dnt]

these should be pretty simple but for some reason i can't get them.

ok they give you a basic time vs distance table:

• 0 sec - 0 m
• 1 sec - 2 m
• 2 sec - 8 m
• 3 sec - 18 m
• 4 sec - 32 m
• 5 sec - 50 m

its a ball rolling down a hill which makes sense for the data. now the question is what is the distance after 2.2 seconds. can you figure that out exactly or do you just look at a graph and estimate it?

second question is what is the slope at 3 seconds? now i konw slope is change in y over change in x (meaning change in distance over change in time) which is also the velocity at 3 seconds but which two points do you use? do you use the 3 second data with the 2 second data or the 3 second data with the 4 second data? they give different answers.

third, if you do look at all the differences from one second to the next, you get a patten of how much it jumps - 2, 6, 10, 14, 18 - those are the velocities at each second, right? well since it jumps by 4 each time can i then say the acceleration is 4 m/s^2?

thanks.

 

[amcavoy]

s = .5at² + v₀t + s₀

Plug in!

 

[dnt]

ok so i can do this:

s = .5(4)(2.2)^2 + (0)(2.2) + 0

is that correct?

am i right the a=4?

 

[amcavoy]

Yes, because 2=.5a -> a=4 m/s².

 

[dnt]

these should be pretty simple but for some reason i can't get them.

ok they give you a basic time vs distance table:

• 0 sec - 0 m
• 1 sec - 2 m
• 2 sec - 8 m
• 3 sec - 18 m
• 4 sec - 32 m
• 5 sec - 50 m

second question is what is the slope at 3 seconds? now i konw slope is change in y over change in x (meaning change in distance over change in time) which is also the velocity at 3 seconds but which two points do you use? do you use the 3 second data with the 2 second data or the 3 second data with the 4 second data? they give different answers.
.

anyone know this question?

 

[mukundpa]

v_{f} = v_{i} + at

 

[dnt]

i don't think that answers my question.

the question specifically asked for slope (which in turn, is velocity) at 3 seconds. to find slope you need two points, right? well do you use 2 and 3 seconds OR 3 and 4 seconds to find the slope at that point?

 

[HallsofIvy]

You are asked, apparently, for an "instantaneous" speed and you don't have enough information to do that. To find instantaneous speed (unless the distance-time graph is a line and speed is constant), you would have to know the distance-time function and you only have it at specific times. What you can do is anyone of what you are suggesting- use the distances at 3 and 4 seconds to get the "average speed between t=3 and t= 4", (32-18)/1= 14 m/s, use the distances at 2 and 3 seconds to get the "average speed between t= 2 and t= 3,(18-8)/1= 10 m/s, or even average those two answers- (10+ 14)/2= 12. For this particular problem, that turns out to be the same as if you find the "average speed between t= 2 and t= 4", symmetric about t= 3, (38- 8)/2= 24/2= 12 m/s. If you want to do a lot of work to get a very accurate answer, you might find a polynomial function that fits all of the points and differentiate that! Actually, here, it is not a "lot of work". It look pretty obvious that each distance, in m, is equal to twice the time, in s, squared: d= 2t².
The instantaneous speed, at t= 3, is the derivative of that, 4t, evaluated at t= 3: 4(3)= 12 m/s.
Of course, that was to be expected. Something moving under gravity has a constant acceleration and so its distance function is a quadratic. That was what apmcavoy was telling you in the first reply to your post.

 

[dnt]

ah. thanks!

 

[curly_ebhc]

Solving the slope

I don't think that you need to take the derivative especially if you are not in a calculus based physics course.
To find the slope at one point draw a tangent line. I promise there is an explantion of how to do this in your book. It should just touch the curve. The tangent line is straight so take any 2 points and find the slope.