What is the acceleration at the 5 s mark?

  • Thread starter Thread starter aatari
  • Start date Start date
  • Tags Tags
    Time Velocity
AI Thread Summary
The discussion revolves around calculating the acceleration of a sports car based on its velocity graph. At the 5-second mark, the car's speed is debated, with initial claims of 4.5 m/s corrected to 3.5 m/s. The acceleration at this point is recalculated to be 0.7 m/s², down from an earlier estimate of 0.9 m/s². For the 15-second mark, participants suggest using two points to determine the slope for acceleration estimation. The conversation also clarifies intervals of negative acceleration, with participants refining their answers based on feedback.
aatari
Messages
73
Reaction score
3
Hey guys I have attempted to solve the question below and hoping to get some feedback. Also, any pointers to solve question 'c' would be really appreciated.

1. Homework Statement

1. A sports car's velocity is shown in the graph below. (Please see the image)

Capture.PNG

a. How fast is it traveling at the 5 s mark?
b. What is the acceleration at the 5 s mark?
c. Estimate the acceleration at the 15 s mark. Show your work.
d. During which intervals of time, did the car have a negative acceleration?​

Homework Equations


Acceleration = Δv/Δt

The Attempt at a Solution


[/B]
a. 4.5 s
b. 0.9 m/s2
c. I can't seem to understand this. Not sure what is the initial velocity in this case. Any help would be appreciated.
d. 11, 12, 13, 19, 20
 
Physics news on Phys.org
For c) what is acceleration?

For d) your answers are not time intervals.
 
Acceleration will be the slope of the line at any point.

a. Are you sure about 4.5?
b. There appears to be a slight curve in the plot from v=0 to v=5. But it is very slight. So if I assume it is straight from t=0, v=0 to t=7.2, v=5, then I get an acceleration of about 0.7. I'm not sure how you got something as high as 0.9.
c. Same thing as step b. Pick 2 points around the 15s mark that appear to be a straight line that closely represent the slope of the curve at the 15s mark and find the slope of that line.
d. It is asking for time intervals where the acceleration is negative. So, for example, you may say that from t=1 to t=4 the car has a negative acceleration (although what I just wrote is not true.)

Edit: I think I just realized how you got 0.9. I think you got it from the 4.5 answer in part a.
 
aatari said:
. 4.5 s
Check this again. Also, the unit should be m/s.
aatari said:
0.9 m/s2
This changes too, because of a).
 
Thanks a lot guys. I double checked my work and here are my answers:

a. 3.5 m/s
b. 3.5/5 = 0.7 m/ss
c. I picked two points: 7.25 - 5.25/15-14 = 2 m/s2
d. t=20.5. t=11, t=12, t=18.5, t=19, t=20
 
aatari said:
Hey guys I have attempted to solve the question below and hoping to get some feedback. Also, any pointers to solve question 'c' would be really appreciated.

1. Homework Statement

1. A sports car's velocity is shown in the graph below. (Please see the image)

View attachment 113473

a. How fast is it traveling at the 5 s mark?
b. What is the acceleration at the 5 s mark?
c. Estimate the acceleration at the 15 s mark. Show your work.
d. During which intervals of time, did the car have a negative acceleration?​

Homework Equations


Acceleration = Δv/Δt

The Attempt at a Solution


[/B]
a. 4.5 s
b. 0.9 m/s2
c. I can't seem to understand this. Not sure what is the initial velocity in this case. Any help would be appreciated.
d. 11, 12, 13, 19, 20
If this is a calculus based course what I would kind of do is this...
fixed.jpg


Now what you have is a series of straight lines instead of the smooth curves. You should now be able to find the equations for those lines, (i.e. to about 9 seconds the equation is just roughly ##v=\frac{2}{3}t##) take the derivative to find the acceleration. For example, at 15 seconds, you have a specific curve that should be easy to find the equation of using any of the techniques in algebra. Take the derivative and you'll find the acceleration for any time on that curve. Your approximate lines should be accurate enough for an estimation.

Alternatively, it might help to plot the acceleration graph on the same graph.

Do this make sense? I know I didn't draw it very well...

Whoops, others already replied..
 
aatari said:
Thanks a lot guys. I double checked my work and here are my answers:

a. 3.5 m/s
b. 3.5/5 = 0.7 m/ss
c. I picked two points: 7.25 - 5.25/15-14 = 2 m/s2
d. t=20.5. t=11, t=12, t=18.5, t=19, t=20

Did you see PeroK's comment on D?
 
Student100 said:
Did you see PeroK's comment on D?
Oh I see. So it will be t = 10.5 to t=12 and t=18.5 to t=20?
 
aatari said:
Oh I see. So it will be t = 10.5 to t=12 and t=18.5 to t=20?

Looks about right.
 
  • #10
Student100 said:
Looks about right.
what about my answers for a, b, and c..are they correct?
 
  • #11
aatari said:
what about my answers for a, b, and c..are they correct?

Looks good enough to me.
 
  • Like
Likes aatari
  • #12
aatari said:
Oh I see. So it will be t = 10.5 to t=12 and t=18.5 to t=20?
If it was me, I would say 10.5 to 13 and 18 to 20. The decrease from 12 to 13 is pretty obvious.
 
  • Like
Likes aatari
  • #13
Thanks a lot everyone for the feedback.
 
Back
Top