<span>Assuming the car is travelling in the same direction for the entire hour, the acceleration is zero.</span>
Answer:
A)![d=\dfrac{1}{2F}mv^2](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7B1%7D%7B2F%7Dmv%5E2)
B)![\Delta KE'=2\times \dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5CDelta%20KE%27%3D2%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
Explanation:
Given that
Force = F
Increase in Kinetic energy = ![\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![\Delta KE=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5CDelta%20KE%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
we know that
Work done by all the forces =change in the kinetic energy
a)
Lets distance = d
We know work done by force F
W= F .d
F.d=ΔKE
![F.d=\dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=F.d%3D%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
![d=\dfrac{1}{2F}mv^2](https://tex.z-dn.net/?f=d%3D%5Cdfrac%7B1%7D%7B2F%7Dmv%5E2)
b)
If the force become twice
F' = 2 F
F'.d=ΔKE'
2 F .d = ΔKE' ( F.d =Δ KE)
2ΔKE = ΔKE'
![\Delta KE'=2\times \dfrac{1}{2}mv^2](https://tex.z-dn.net/?f=%5CDelta%20KE%27%3D2%5Ctimes%20%5Cdfrac%7B1%7D%7B2%7Dmv%5E2)
Therefore the final kinetic energy will become the twice if the force become twice.
Answer:
![v_{max}=52.38\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D52.38%5Cfrac%7Bm%7D%7Bs%7D)
![v_{100}=33.81](https://tex.z-dn.net/?f=v_%7B100%7D%3D33.81)
Explanation:
the maximum speed is reached when the drag force and the weight are at equilibrium, therefore:
![\sum{F}=0=F_d-W](https://tex.z-dn.net/?f=%5Csum%7BF%7D%3D0%3DF_d-W)
![F_d=W](https://tex.z-dn.net/?f=F_d%3DW)
![kv_{max}^2=m*g](https://tex.z-dn.net/?f=kv_%7Bmax%7D%5E2%3Dm%2Ag)
![v_{max}=\sqrt{\frac{m*g}{k}} =\sqrt{\frac{70*9.8}{0.25}}=52.38\frac{m}{s}](https://tex.z-dn.net/?f=v_%7Bmax%7D%3D%5Csqrt%7B%5Cfrac%7Bm%2Ag%7D%7Bk%7D%7D%20%3D%5Csqrt%7B%5Cfrac%7B70%2A9.8%7D%7B0.25%7D%7D%3D52.38%5Cfrac%7Bm%7D%7Bs%7D)
To calculate the velocity after 100 meters, we can no longer assume equilibrium, therefore:
![\sum{F}=ma=W-F_d](https://tex.z-dn.net/?f=%5Csum%7BF%7D%3Dma%3DW-F_d)
![ma=W-F_d](https://tex.z-dn.net/?f=ma%3DW-F_d)
![ma=mg-kv_{100}^2](https://tex.z-dn.net/?f=ma%3Dmg-kv_%7B100%7D%5E2)
(1)
consider the next equation of motion:
![a = \frac{(v_{x}-v_0)^2}{2x}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%28v_%7Bx%7D-v_0%29%5E2%7D%7B2x%7D)
If assuming initial velocity=0:
(2)
joining (1) and (2):
![\frac{v_{100}^2}{2x}=g-\frac{kv_{100}^2}{m}](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7B100%7D%5E2%7D%7B2x%7D%3Dg-%5Cfrac%7Bkv_%7B100%7D%5E2%7D%7Bm%7D)
![\frac{v_{100}^2}{2x}+\frac{kv_{100}^2}{m}=g](https://tex.z-dn.net/?f=%5Cfrac%7Bv_%7B100%7D%5E2%7D%7B2x%7D%2B%5Cfrac%7Bkv_%7B100%7D%5E2%7D%7Bm%7D%3Dg)
![v_{100}^2(\frac{1}{2x}+\frac{k}{m})=g](https://tex.z-dn.net/?f=v_%7B100%7D%5E2%28%5Cfrac%7B1%7D%7B2x%7D%2B%5Cfrac%7Bk%7D%7Bm%7D%29%3Dg)
![v_{100}^2=\frac{g}{(\frac{1}{2x}+\frac{k}{m})}](https://tex.z-dn.net/?f=v_%7B100%7D%5E2%3D%5Cfrac%7Bg%7D%7B%28%5Cfrac%7B1%7D%7B2x%7D%2B%5Cfrac%7Bk%7D%7Bm%7D%29%7D)
(3)
![v_{100}=\sqrt{\frac{9.8}{(\frac{1}{2*100}+\frac{0.25}{70})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B1%7D%7B2%2A100%7D%2B%5Cfrac%7B0.25%7D%7B70%7D%29%7D%7D)
![v_{100}=\sqrt{\frac{9.8}{(\frac{1}{200}+\frac{1}{280})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B1%7D%7B200%7D%2B%5Cfrac%7B1%7D%7B280%7D%29%7D%7D)
![v_{100}=\sqrt{\frac{9.8}{(\frac{3}{350})}}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B%5Cfrac%7B9.8%7D%7B%28%5Cfrac%7B3%7D%7B350%7D%29%7D%7D)
![v_{100}=\sqrt{1,143.3}](https://tex.z-dn.net/?f=v_%7B100%7D%3D%5Csqrt%7B1%2C143.3%7D)
![v_{100}=33.81](https://tex.z-dn.net/?f=v_%7B100%7D%3D33.81)
To plot velocity as a function of distance, just plot equation (3).
To plot velocity as a function of time, you have to consider the next equation of motion:
![v = v_0 +at](https://tex.z-dn.net/?f=v%20%3D%20v_0%20%2Bat)
as stated before, the initial velocity is 0:
(4)
joining (1) and (4) and reducing you will get:
![\frac{kt}{m}v^2+v-gt=0](https://tex.z-dn.net/?f=%5Cfrac%7Bkt%7D%7Bm%7Dv%5E2%2Bv-gt%3D0)
solving for v:
![v=\frac{ \sqrt{1+\frac{4gk}{m}t^2}-1}{\frac{2kt}{m} }](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B%20%5Csqrt%7B1%2B%5Cfrac%7B4gk%7D%7Bm%7Dt%5E2%7D-1%7D%7B%5Cfrac%7B2kt%7D%7Bm%7D%20%7D)
Plots:
Anger increases until an emotional breakdown happens.
Answer:
B) weather conditions that cannot be foreseen, and pilot errors often cause much larger deposits of spray than
Explanation:
Option B obeys the principle of parallelism because the two phrases, " can be carefully planned" and "cannot be foreseen" follow the same grammatical structure. Option A does not obey parallelism because " can be carefully planned" and "could not be foreseen" do not follow the same grammatical structure.
The phrase " they had anticipated" in the other options apart from B do not have a matching subject. Pilot errors is the subject in this sentence and it is not expected to anticipate anything, only the pilots can.
Moreover, "that cannot be foreseen" is more parallel to "can be carefully planned" that "unforeseeable" or "foreseeable" used in the other options