Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
The maximum force acts between B and C as the graph is steepest showing maximum deceleration
Answer:
i don't know if this is good for you but
Explanation:
ignoring frictional air resistance (drag) the speed on return is the same as when it left the ground (5 m/s but in the opposite direction).
Note: this points out a good reason for not firing live bullets into the air..they will return somewhere and at the same speed.
However, if you take into account the atmospheric drag the reurn speed will be somewhat smaller (but in the case of a bullet, probably still lethal.) Drag depends on many factors and is difficult to calculate.
control is constant for given set of readings.
set indep var
observe dep var
It causes the global pattern of wind and precipitation (such as rain, snow, or hail).
