# Questions on kinematics (distance/velocity/acceleration)

these should be pretty simple but for some reason i cant 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.

s = .5at2 + v0t + s0

Plug in!

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?

Yes, because 2=.5a -> a=4 m/s2.

dnt said:
these should be pretty simple but for some reason i cant 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
Homework Helper
$$v_{f} = v_{i} + at$$

i dont 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
Homework Helper
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.

ah. thanks!

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.

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