Amotorcyclist 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?

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

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.