How to Calculate Acceleration for a Stopping Sports Car

  • Thread starter thschica
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In summary, the sports car advertised has a deceleration of -0.711 m/s2 or -0.073 g's when stopping from a speed of 85 km/h to 0 in a distance of 55 m. This information can be found using the kinematics equation v^2 - u^2 = 2as, where v_f = 0 m/s, v_i = 23.611 m/s, and d = 55 m.
  • #1
thschica
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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!
 
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  • #2
Do you know the equation:
[tex]v_f ^ 2 = v_i ^ 2 + 2ad[/tex]?
It's used to solve for a when v_i, v_f, and d is known.
Viet Dao,
 
  • #3
would v_i =23.6 m/s,v_f=55m,and d=0
 
  • #4
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?
 
  • #5
I have no idea what to do in this problem.I divided 85km/h to get it into m/s.
 
  • #6
thschica said:
would v_i =23.6 m/s,v_f=55m,and d=0

[tex] v_f = 0[/tex] m/s
[tex] v_i = 23.611[/tex] m/s
[tex]d = 55[/tex] m
 
  • #7
Use the basic kinematics equation
[itex]v^2 - u^2 = 2as[/itex]

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
 
  • #8
Thank you!I understand what I did!Is the answer a negative number?
 
  • #9
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.
 

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