Q in instantanous acceleration I want check my answer .

Also, it would be clearer if you mentioned that the points used were (1.4, 1.5) and (1.6, 3.0), as this is not explicitly stated in the question.In summary, the conversation discusses instantaneous acceleration and how it is calculated using the average acceleration formula. The correctness of the answer provided by the person is questioned and improvements are suggested for parts a) and c) of the solution. The person is also reminded to state the coordinates of the points used in the calculation.
  • #1
r-soy
172
1
Hi all

1.JPG


Q in instantaneous acceleration ?



https://www.physicsforums.com/attachments/17036



my answer is :



a ) As we know Instanteouse acceleration :

It is the limit of the average acceleration when the time interval to zero



a = changing in v / changing t
3.0-1.5 / 0.8 - 0.4 = 35225 m/s



b )

instantanous acceleration = Zero because it's straight line



c ) here the gragh go down then the ( a ) will by ( - )



a = changing in v / changing t



3.0 - 1.5 / 1.6 - 1.4 = -0.4625



this is my answer I want check
 
Physics news on Phys.org
  • #2
Using your method part a) comes to 1.5/0.4=3.75m/s^2.Your method is right but something went wrong with your maths(used calculator perhaps and pressed wrong buttons?).Please note that you could have used the whole triangle and written a=3/0.8.

You made a similar mistake in part c)

You need to improve on your reasoning for part b).The acceleration is zero because it is a straight line parallel to the time axis,in other words there is no change of velocity.
 
  • #3
hi thanks

but what are the mistake in part c ?
 
  • #4
You wrote a=3-1.5/1.6-1.4.It would be clearer if you wrote a=-(3-1.5)/(1.6-1.4)=1.5/0.2=-7.5m/s^2
 
  • #5
my answer..



I would first like to commend you for your effort in solving this problem and checking your answer. It is important to always double-check our work in science to ensure accuracy. Your method for calculating instantaneous acceleration is correct, as it involves taking the limit of the average acceleration as the time interval approaches zero. However, I would suggest using the correct units for acceleration, which is meters per second squared (m/s^2) instead of meters per second (m/s). Additionally, it is important to note that instantaneous acceleration can have a non-zero value even on a straight line, depending on the velocity and time values at a specific point. Therefore, it is not always equal to zero. Lastly, when calculating acceleration using the formula a = (vf - vi) / (tf - ti), it is important to use the correct velocity values at the specific point you are interested in, rather than the overall change in velocity. Overall, your approach is correct, but it is always good to double-check and make sure you are using the correct units and values in your calculations. Keep up the good work!
 

1. What is instantaneous acceleration?

Instantaneous acceleration is the rate at which an object's velocity changes at a specific moment in time. It is the slope of the velocity vs. time graph at a given point.

2. How is instantaneous acceleration different from average acceleration?

Instantaneous acceleration is the acceleration at a specific point in time, while average acceleration is the average rate of change in velocity over a given time interval.

3. Can you give an example of instantaneous acceleration?

Yes, an example of instantaneous acceleration is when a car is going around a curve on a race track. The car's velocity is constantly changing, and at any given point on the curve, the instantaneous acceleration can be calculated.

4. How is instantaneous acceleration calculated?

Instantaneous acceleration can be calculated by taking the derivative of an object's velocity function with respect to time.

5. Why is instantaneous acceleration important in physics?

Instantaneous acceleration is important because it allows us to analyze an object's motion at a specific point in time, rather than just looking at its overall average acceleration. It also helps us understand how forces affect an object's velocity and position at different points in time.

Similar threads

  • Introductory Physics Homework Help
Replies
5
Views
978
  • Introductory Physics Homework Help
Replies
5
Views
774
  • Introductory Physics Homework Help
Replies
10
Views
12K
  • Introductory Physics Homework Help
Replies
6
Views
6K
  • Introductory Physics Homework Help
Replies
3
Views
962
  • Introductory Physics Homework Help
Replies
15
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
613
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
5K
  • Introductory Physics Homework Help
Replies
1
Views
3K
Back
Top