A motorcyclist traveling at 30.7 km/h accelerates after passing a sign post marking the city limits. If it takes 4.9 seconds for the motorcyclist to reach 120 km/h, how far has he traveled (in meters) past the sign post?
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Remark
Writing down the givens for these questions is almost a command. You must know which equation of the big 4 to use. The givens + what you want to find will determine the equation.
Givens
vi = 30.7 km/hr = 30.7 km/hr *[1000m/1km]*[1hr/3600sec] = 8.52m/sec
vf = 120 km/hr = 120 km/hr * [1000 m/1 km]*[1h/3600 sec] = 33.3 m/s
t = 4.9 seconds
d = ???
Formula
d = (vi + vf)*t/2
Solve
d = ( 33.3+8.52 )*4.9/2
d = (41.85333)*4.9/2
d = 205.02/2
d = 102.51 m<<<Answer
The average speed over the interval is
... (30.7 km/h + 120 km/h)/2 = 75.35 km/h
In 4.9 seconds, the corresponding distance is
... (75.35 km/h)*(4.9/3600 h)*(1000 m/km) = 103 m