You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an alternative browser.
- Forums
- Homework Help
- Introductory Physics Homework Help
Sports Car Acceleration Question
- Thread starter Adrianna
- Start date Jan 14, 2010
Homework Statement
A sports car moving at constant speed travels 110m in 5.0s. If it then brakes and comes to a stop in 4.0s. what is its acceleration in m/s^2? Express the answer in terms of "g's." where 1.00 g = 9.80 m/s^2. Homework Equations
I thought about using the equation a=v/tThe Attempt at a Solution
So I tried it sort of I am just
sort of lost I am not really sure how to start....
Answers and Replies
- Jan 14, 2010
- #2
Homework Equations
I thought about using the equation a=v/t
You need to use the kinematic equations
v=u+at
v2=u2+2as
s=ut+1/2at2
Which one do you think you will have to use?
- Jan 14, 2010
- #3
What does the u stand for, that's not in my book?
- Jan 14, 2010
- #4
The car initially had a velocity of 110m/5s (22m/s) and came to a stop (0m/s) in 4.0s. The acceleration is the change in velocity over a time interval.
- Jan 14, 2010
- #5
So would we use the v=u+at since I have the 4.0s and the 22m/s? but what would go in for the u?
- Jan 14, 2010
- #6
u is initial velocity and v is final velocity.
- Jan 14, 2010
- #7
Okay I figured it out thanks so much the answer was -5.5 m/s^2 and then in g's it was -0.56 g's
- Jan 14, 2010
- #8
take u (initial speed) = 110 / 5 = 22m/s v = u + at
v = 0
t = 4
0 = 22 + 4a
4a = -22
a = -5.5m/s^2
a = (-5.5/9.8)g = -0.56g
Suggested for: Sports Car Acceleration Question
- Last Post
- Oct 6, 2007
- Last Post
- Apr 25, 2012
- Last Post
- Aug 26, 2015
- Last Post
- Oct 31, 2006
- Last Post
- Sep 1, 2015
- Last Post
- Jan 26, 2005
- Last Post
- Sep 3, 2017
- Last Post
- Oct 13, 2008
- Last Post
- Jan 29, 2006
- Last Post
- Sep 2, 2021
- Forums
- Homework Help
- Introductory Physics Homework Help
Pau L. a. How fast is the far moving during the first 5 s?
b. What is the magnitude of the acceleration of the car as it brakes and comes to a stop?
More
1 Expert Answer
a. You are given distance and time, and velocity is the ratio of the two.
b. The car goes from a velocity of V1 to velocity of V2 in a given time. Acceleration is the rate of change of velocity, so, a = (V1 - V2) / (t1 - t2)
Note: This is a very simple question that for AP Physics should have been well demonstrated in the textbook. Reread the textbook and your class notes to see where this is explained.
Still looking for help? Get the right answer, fast.
OR
Find an Online Tutor Now
Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.