How to Calculate Acceleration for a Stopping Sports Car

  • Thread starter Thread starter thschica
  • Start date Start date
AI Thread Summary
To calculate the acceleration of a sports car stopping from 85 km/h over a distance of 55 m, the kinematic equation v_f^2 = v_i^2 + 2ad can be used, where v_f is 0 m/s, v_i is 23.611 m/s, and d is 55 m. The resulting acceleration will be negative, indicating deceleration. After calculating the acceleration in m/s², it can be converted to g's by dividing by 9.80 m/s². The discussion clarifies the importance of converting units and understanding the signs of acceleration. Negative acceleration signifies the car is slowing down.
thschica
Messages
47
Reaction score
0
A sports car is advertised to be able to stop in a distance of 55 m from a speed of 85 km/h.
(a) What is its acceleration in m/s2?

(b) How many g's is this (g = 9.80 m/s2)?




I am confused as to how you would find the acceleration.Iwould be able to put it into g's but i don't understand how to get the acceleration!
 
Last edited:
Physics news on Phys.org
Do you know the equation:
v_f ^ 2 = v_i ^ 2 + 2ad?
It's used to solve for a when v_i, v_f, and d is known.
Viet Dao,
 
would v_i =23.6 m/s,v_f=55m,and d=0
 
thschica said:
would v_i =23.6 m/s,v_f=55m,and d=0

No. Where do you get that 23.6 ms^-1 from? v_f is also a speed, so it isn't a distance. Also, why did you put d = 0?
 
I have no idea what to do in this problem.I divided 85km/h to get it into m/s.
 
thschica said:
would v_i =23.6 m/s,v_f=55m,and d=0

v_f = 0 m/s
v_i = 23.611 m/s
d = 55 m
 
Use the basic kinematics equation
v^2 - u^2 = 2as

For your second part , divide your answer by g , that will give you the reqd. number.

Donot forget to convert km/hr to m/sec.

Regards
BJ
 
Thank you!I understand what I did!Is the answer a negative number?
 
thschica said:
Thank you!I understand what I did!Is the answer a negative number?

Yes, indeed. The velocity has decreased down to zero which means that the acceleration must be negative. A negative acceleration is also called deceleration/retardation.
 
Back
Top