A racing car undergoing constant acceleration covers 160 m in 3.7secs. (a) If it

Question

A racing car undergoing constant acceleration covers 160 m in 3.7secs.
  (a) If it is moving at 55 m/s at the end of this interval, whatwas its speed at the beginning of the interval?
this is what I got:
avg V= (160/3.7)
using formula avg V = 1/2 (v1+v2) i can get v1since I know avg V and V2
   V1= 31 m/s.
(b) How far did it travel FROM REST to the end of the160 m distance?
This is where I don't know how to continue, I can't find aformula for this problem.

  (a) If it is moving at 55 m/s at the end of this interval, whatwas its speed at the beginning of the interval?
this is what I got:
avg V= (160/3.7)
using formula avg V = 1/2 (v1+v2) i can get v1since I know avg V and V2
   V1= 31 m/s.
(b) How far did it travel FROM REST to the end of the160 m distance?
This is where I don't know how to continue, I can't find aformula for this problem.
  (a) If it is moving at 55 m/s at the end of this interval, whatwas its speed at the beginning of the interval?
this is what I got:
avg V= (160/3.7)
using formula avg V = 1/2 (v1+v2) i can get v1since I know avg V and V2
   V1= 31 m/s.
(b) How far did it travel FROM REST to the end of the160 m distance?
This is where I don't know how to continue, I can't find aformula for this problem.

Explanation / Answer

(a)The final speed of the car is v = u + gt or u = v - gt v = 55 m/s,g = 9.8 m/s2 and t = 3.7 s or u = 55 - 9.8 * 3.7 = 18.74 m/s (b)The acceleration of the car at the end of 160 m distanceis S = ut + (1/2)a * t2 or a = [2 * (S - ut)/t2] or a = [2 * (160 - 0 * 3.7)/(3.7)2] = 23.3m/s2 we know that v2 - u2 = 2aS or S = (v2 - u2/2a) or S = ((55)2 - 02/2 * 23.3) = 64.9m we know that v2 - u2 = 2aS or S = (v2 - u2/2a) or S = ((55)2 - 02/2 * 23.3) = 64.9m

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