Question

TheJamaicanbobsledteamhitthebrakesontheir sledsothatitdeceleratesatauniformrateof0.43 m/s².Howlongdoesittaketostopifittravels85m beforecomingtorest?

Answer 1

Explanation:

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