See the attached picture:
Answer:
largest lead = 3 m
Explanation:
Basically, this problem is about what is the largest possible distance anchorman for team B can have over the anchorman for team A when the final leg started that anchorman for team A won the race. This show that anchorman for team A must have higher velocity than anchorman for team B to won the race as at the starting of final leg team B runner leads the team A runner.
So, first we need to calculate the velocities of both the anchorman
given data:
Distance = d = 100 m
Time arrival for A = 9.8 s
Time arrival for B = 10.1 s
Velocity of anchorman A = D / Time arrival for A
=100/ 9.8 = 10.2 m/s
Velocity of anchorman B = D / Time arrival for B
=100/10.1 = 9.9 m/s
As speed of anchorman A is greater than anchorman B. So, anchorman A complete the race first than anchorman B. So, anchorman B covered lower distance than anchorman A. So to calculate the covered distance during time 9.8 s for B runner, we use
d = vt
= 9.9 x 9.8 = 97 m
So, during the same time interval, anchorman A covered 100 m distance which is greater than anchorman B distance which is 97 m.
largest lead = 100 - 97 = 3 m
So if his lead no more than 3 m anchorman A win the race.
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash
ENERGY = POWER X TIME
=60 X 120=7200KWh
