How Do I Solve These Car Acceleration Problems?

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Homework Help Overview

The discussion revolves around car acceleration problems, specifically focusing on calculating time and speed given initial and final velocities along with acceleration. The subject area includes kinematics and basic physics principles related to motion.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants explore the relationship between acceleration, velocity, and time, questioning the correct application of formulas. Some participants attempt to clarify the calculations for determining time and speed based on given acceleration.

Discussion Status

The discussion is ongoing, with participants providing guidance and corrections to each other's calculations. Multiple interpretations of the problems are being explored, and there is a collaborative effort to clarify misunderstandings without reaching a definitive conclusion.

Contextual Notes

Participants express difficulties with the formulas and calculations, indicating a need for clearer understanding of the concepts involved. There are mentions of formatting issues that affect the clarity of the presented solutions.

nina09
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Hi , I'm new to to the forum . I have a very hard time with physics or maybe is undestanding the professor . I currently trying to solve this acceleration problems but I end up not understanding that I need to do next or how to solve them. If any you can help me I will really apreaciate it.


1)The tires of a car begin to lose their grip on the road at an acceleration of 5 m/s2. At this accelaration , how long does the car need to reach speed of 25m/s starting from10 m/s?


A=5 m/s2
V1= 10 m/s
V2=25 m/s




T= 10m/s
------ = 2s
5m/s2



2) A car start from rest and reaches a speed of 40m/s in 10s. If its aceleration remain the same how fast will it be moving 5s later?

V1=0
V2=40m/s
T=10s

A= 40m/s -0 40 ms
------------ =--------
10s 10s

 
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Welcome to PF!

Hi nina09! Welcome to PF! :smile:
nina09 said:
1)The tires of a car begin to lose their grip on the road at an acceleration of 5 m/s2. At this accelaration , how long does the car need to reach speed of 25m/s starting from10 m/s?


A=5 m/s2
V1= 10 m/s
V2=25 m/s




T= 10m/s
------ = 2s
5m/s2

No.

Acceleration is rate of change of velocity …

so how much is the velocity changing? and how long does it take to change? :wink:
2) A car start from rest and reaches a speed of 40m/s in 10s. If its aceleration remain the same how fast will it be moving 5s later?

V1=0
V2=40m/s
T=10s

A= 40m/s -0 40 ms
------------ =--------
10s 10s

Yes … keep going! :smile:
 
Hi Nina and welcome to PF!
It is very natural to have trouble with any new skill at first; you just have to work hard at it and it may turn out that you have a lot of talent.
A=5 m/s2
V1= 10 m/s
V2=25 m/s
is a good start. Look in your list of formulas for one that relates the velocities, acceleration and time. The one I'm thinking of is
a = Δv/Δt. The Δ on the t doesn't make any difference since we start at time zero, but the Δ on the v does because you start at v=10. So you get Δt = Δv/a = (25-10)/5

In the second problem, it is difficult to follow your calc because this forum removes spaces. I think you wrote
a = Δv/Δt = 40/10 = 4 m/s^2.
This is correct and knowing the acceleration you should be able to use the same formula again over 15 seconds to find the velocity.
 
I wrote each step under the othe so is easier to see the results

This is my solution to the second one , Did I did it correctly?
1) 40m/s -0
-------------=
15s

2)40m/s
-------- =
15s

3) 2.6m/s^2
------------
15 s


I'm still having dificulty's with the first one since I can't find the formula.
 
For the second one, it should be
a = Δv/Δt = 40/10 = 4 m/s^2
The car goes from 0 to 40 m/s in 10 seconds, not 15.
In a second step you use the "same acceleration" of 4 with a time of 15 to get the speed after 15 seconds.

In the first one, the formula is a = Δv/Δt. Your only mistake was to say Δv = 10 when it was actually 25-10.
 
This my second try at the first one, is it wrong?

25m/s-10ms
5m/s^2= -------------=
T

15m/s x 5m/s^2
------------ =
T

=0.6m/s^2
 
First one should be Δt = Δv/a = (25-10)/5 = 3 m/s^2.
 
This is what I understood from the explanation for the second one:

40m/s- 0
---------=
10s

40m/s
------- = 4m/s^2
10s

2 step

4m/s^2
--------
15s

=1m/s^2
 
First one

25m/s-10m/s
-------------=
5m/s^2


25m/s-10m/s
-------------= 3m/s^2
5m/s^2
 
  • #10
First one looks good!
Second one, second step should be
a = Δv/Δt
a*Δt = Δv after multiplying both sides by Δt
Δv = a*Δt = . . .
 
  • #11
Does this * means multiplication?
 
  • #12
Yes, * means multiply. If you don't use some *s, things can get unreadable.
 
  • #13
This the second one, I know I'm doing something wrong but I just can't pinpoint where exactly I'm making the mistake.


40m/s- 0
---------=
10s

40m/s
------- = 4m/s^2
10s


Second step

The acceleration is 4m/s^2 and the time is 15


A*t
-----=
V


4m/s^2 *15s = 60m/s
------------
V

60m/s
------- =4
15s

?
 
  • #14
I said a = Δv/Δt
Use a=4, Δt = 15 and this is
4 = Δv/15
Multiply both sides by 15 to get the Δv by itself:
15*4 = Δv
60 = Δv

Will try writing this in your notation. Have to use dots to make it line up.
...Δv
4 = ----
...15

...Δv * 15
15*4 = ----
...15

...Δv
15*4 = ----
...1
60 = Δv
 
  • #15
I think I got it , thank you very much.


40m/s- 0
---------=
--10s

40m/s
------- = 4m/s^2
--10s
------V
4 = ----
------15

-------Vx15
15*4 = ----
-------15

----------V
15*4 = ----
--------- 1

60 = V
 
  • #16
I love that trick with the white dashes! Thank you!
Another good trick is, if you need a symbol like Δ or θ, you just go to
https://www.physicsforums.com/blog.php?b=347
and copy it.
 
Last edited by a moderator:
  • #17
Thank you .
 

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