Answer:
B meet A 0.01 km east of flagpole
Explanation:
given data
distance A = 5.7 km west
velocity V1 = 8.9 km/h
distance B = 4.5 km east
velocity V2 = 7 km/h
to find out
How far runners from the flagpole, when paths cross
solution
we know A and B are 5.7 + 4.5 = 10.2 km apart
and we consider here B will run distance x km for meet
so time will be for B is
time B = distance / velocity
time B = x / 7 ...................1
and
for A distance for meet = ( 10.2 - x ) km
so time A = distance / velocity
time A = ( 10.2 - x ) / 8.9 .............2
now equating equation 1 and 2
time A = time B
x / 7 = ( 10.2 - x ) / 8.9
x = 4.490
so distance of B run for meet is 4.490 km
so distance from the flagpole when their paths cross is 4.5 - 4.490 = 0.01 km
so B meet A 0.01 km east of flagpole
Answer:
Explanation:
Question is incomplete
Assuming the question you have asked is
You are driving home from school steadily at 95 km/h for 180 km. It then begins to rain and you slow to 65 km/h. You arrive home after driving 4.5 h.
given,
speed of 95 km/h for 180 km
due to rain
speed is reduced to 65 km/h
distance traveled in 4.5 hour
time taken to travel 180 km
d = s x t

t = 1.9 hr
distance traveled in time, t' = 4.5-1.9 = 2.6 hr
Speed of vehicle = 65 Km/h
d' = s x t'
d' = 65 x 2.6
d'= 169 Km
total distance your hometown from school
D = d + d'
D = 180 + 169
D = 349 Km
It'll be 152 Hz at the exact instant the bumblebee
is right at the tip of your nose, on his way past you.
Before he gets there, while he's coming at you,
he sounds like a frequency higher than 152 Hz.
After he passes by, and is going away from you,
he sounds like a frequency lower than 152 Hz.
Answer:
It has been learned in this lesson that the area bounded by the line and the axes of a velocity-time graph is equal to the displacement of an object during that particular time period. ... Once calculated, this area represents the displacement of the object.
Explanation: