Respuesta :
here's a cheap trick
it would take the same time to accelerate from rest to top speed
as it would take to decelerate from top speed to zero
so
instead of
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative
we'll use
Vi = 0 and a is positive
giving
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2
t^2 = 395.345
t = 19.88s or 20. s to 2 sig figs
or we ccould find Vi from
Vf*2 = Vi^2 + 2 a d
0 = Vi^2 + 2 (0.43) 85
Vi^2 = 71.4
Vi = 8.45m/s
then
85 = 8.45 t + 1/2 (-0.43) t^2
85 = 8.45 t - 0.215 t^2
0.215 t^2 - 8.45 + 85 = 0
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi)
or another cheap trick
when a is constant
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225
and
d = Vavg t
85 = 4.225 t
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings)
anyway you choose you get 20. s
it would take the same time to accelerate from rest to top speed
as it would take to decelerate from top speed to zero
so
instead of
d = Vi t + 1/2 a t^2 where Vi is positive and a is negative
we'll use
Vi = 0 and a is positive
giving
85 = 0 + 1/2 (0.43) t^2 = 0.215 t^2
t^2 = 395.345
t = 19.88s or 20. s to 2 sig figs
or we ccould find Vi from
Vf*2 = Vi^2 + 2 a d
0 = Vi^2 + 2 (0.43) 85
Vi^2 = 71.4
Vi = 8.45m/s
then
85 = 8.45 t + 1/2 (-0.43) t^2
85 = 8.45 t - 0.215 t^2
0.215 t^2 - 8.45 + 85 = 0
t = 19.65s or 20. s to 2 s.f.(minor difference arises from rounding Vi)
or another cheap trick
when a is constant
Vavg = (Vf + Vi) /2 = 8.45/2 = 4.225
and
d = Vavg t
85 = 4.225 t
t = 20.12 or 20. s to 2 s.f. (minor differences from intermidiate roundings)
anyway you choose you get 20. s
