How to calculate instantaneous acceleration?

AI Thread Summary
To calculate instantaneous acceleration, one must determine the slope of the velocity-time graph at the specific point of interest. In this case, since the acceleration is uniform, the instantaneous acceleration equals the average acceleration. The slope of the graph from 30 seconds to 40 seconds is constant, indicating uniform acceleration. At 35 seconds, the instantaneous acceleration is calculated to be -1.667 m/s². Understanding the tangent line at the point of interest is crucial for accurate calculation.
treyh
Messages
3
Reaction score
1
This is the graph and question:
http://img193.imageshack.us/img193/3489/physicquestion.jpg

I cannot seem to find the instantaneous acceleration. I must be doing it wrong because I'm pretty sure to find the the inst. acceleration all i need to know is the tangent line equation.

I've tried a bunch of answers but can't seem to get it right. I need to express the answer in m/s^2 and not km/h but that is easy to convert.

I haven't had much physics experience. How do i find the instantaneous acceleration?

Thanks
 
Last edited by a moderator:
  • Like
Likes Tintin Javier
Physics news on Phys.org
Since the acceleration is uniform, instantaneous acceleration = average acceleration.

acceleration = slope of v-t graph

from the graph, the slope of the graph during time = 30s to 40s is a straight line (that means the slope is not varying), that means the acceleration uniform.

so the instantaneous acceleration at time = 35s :
= [(60/3.6) - 0 / (30 - 40)] = -1.667 m/s^2.
 
Last edited:
  • Like
Likes Tintin Javier
treyh said:
I'm pretty sure to find the the inst. acceleration at t = 35 seconds [/color] all i need to know is the tangent line equation.
That's correct, so what is the definition of instantaneous acceleration and what is the slope of that line at t = 35 s?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top