physics
posted by Grace .
A car is traveling at 95km/hr. the driver steps on the brakes and the car comes to a stop in 60m. What is the car's deceleration?

Vi = 95,000m/3600 s = 26.4 m/s
v = Vi  a t
v = 0 at end of trip
0 = 26.4  a t so t = 26.4/a
x = Vi t  .5 a t^2
60 = 26.4 t  .5 a t^2
60 = 26.4 (26.4/a) .5 (a)(26.4^2)/a^2
60 = .5 (26.4)^2/a
a = 5.8 m/s^2 is deacceleration
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