Yes. Displacement is a vector, as opposed to distance, which is a scalar. Since displacement is a vector quantity, the negative sign represents direction. As with velocity and acceleration, it just comes down to how you define your coordinate system
When plane is flying in tail wind condition we will have
![V_{plane} + v_{wind} = \frac{distance}{time}](https://tex.z-dn.net/?f=V_%7Bplane%7D%20%2B%20v_%7Bwind%7D%20%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D)
![150 mph + v_{wind} = \frac{distance}{2}](https://tex.z-dn.net/?f=150%20mph%20%2B%20v_%7Bwind%7D%20%3D%20%5Cfrac%7Bdistance%7D%7B2%7D)
Now during return journey we can say its would be head wind now
![V_{plane} - v_{wind} = \frac{distance}{time}](https://tex.z-dn.net/?f=V_%7Bplane%7D%20-%20v_%7Bwind%7D%20%3D%20%5Cfrac%7Bdistance%7D%7Btime%7D)
time = 2 + 1 = 2 hours
![150 - v_{wind} = \frac{distance}{3}](https://tex.z-dn.net/?f=150%20-%20v_%7Bwind%7D%20%3D%20%5Cfrac%7Bdistance%7D%7B3%7D)
now add two equations
![300 = \frac{5*distance}{6}](https://tex.z-dn.net/?f=300%20%3D%20%5Cfrac%7B5%2Adistance%7D%7B6%7D)
distance = 360 miles
now from above equation again
![v_{wind} = 30 mph](https://tex.z-dn.net/?f=v_%7Bwind%7D%20%3D%2030%20mph)
Motion occurs because a force per it into action
Answer:
gravity
Explanation:
Gravity pulls the planets toward the Sun. Gravity pulls the moon toward Earth. Gravity pulls us toward the Earth. Gravity is a force. Inertia.
The work done is ![2.8125 \times 10^{5} \mathrm{J}](https://tex.z-dn.net/?f=2.8125%20%5Ctimes%2010%5E%7B5%7D%20%5Cmathrm%7BJ%7D)
Work Done = Change in Kinetic Energy (ΔKE)
<u>Explanation</u>
In first 1 hour it travels 72 km
So, Velocity = ![\frac{\text { distance }}{\text { time }}=\frac{72}{1} k m / h=72 \mathrm{km} / \mathrm{h}=\frac{72000}{3600} \mathrm{m} / \mathrm{s}=20 \mathrm{m} / \mathrm{s}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7B%20distance%20%7D%7D%7B%5Ctext%20%7B%20time%20%7D%7D%3D%5Cfrac%7B72%7D%7B1%7D%20k%20m%20%2F%20h%3D72%20%5Cmathrm%7Bkm%7D%20%2F%20%5Cmathrm%7Bh%7D%3D%5Cfrac%7B72000%7D%7B3600%7D%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%3D20%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D)
or, Initial Velocity (u) = 20 m/s
Similarly for the next hour it covers 90 km
So, Velocity = ![\frac{\text { distance }}{\text { time }}=\frac{90}{1} k m / h=90 \mathrm{km} / \mathrm{h}=\frac{90000}{3600} \mathrm{m} / \mathrm{s}=25 \mathrm{m} / \mathrm{s}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Ctext%20%7B%20distance%20%7D%7D%7B%5Ctext%20%7B%20time%20%7D%7D%3D%5Cfrac%7B90%7D%7B1%7D%20k%20m%20%2F%20h%3D90%20%5Cmathrm%7Bkm%7D%20%2F%20%5Cmathrm%7Bh%7D%3D%5Cfrac%7B90000%7D%7B3600%7D%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D%3D25%20%5Cmathrm%7Bm%7D%20%2F%20%5Cmathrm%7Bs%7D)
or, Final Velocity (v) = 20 m/s
Work done = Change in Kinetic Energy (ΔKE)
Work done = ΔKE = ![\frac{1}{2} m v^{2}-\frac{1}{2} m u^{2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%20v%5E%7B2%7D-%5Cfrac%7B1%7D%7B2%7D%20m%20u%5E%7B2%7D)
ΔKE = ![\frac{1}{2} m\left(v^{2}-u^{2}\right)=\frac{1}{2} \times\left(2.5 \times 10^{3}\right) \times\left(25^{2}-20^{2}\right)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m%5Cleft%28v%5E%7B2%7D-u%5E%7B2%7D%5Cright%29%3D%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes%5Cleft%282.5%20%5Ctimes%2010%5E%7B3%7D%5Cright%29%20%5Ctimes%5Cleft%2825%5E%7B2%7D-20%5E%7B2%7D%5Cright%29)
=
= 281250 joule
= ![2.8125 \times 10^{5} \mathrm{J}](https://tex.z-dn.net/?f=2.8125%20%5Ctimes%2010%5E%7B5%7D%20%5Cmathrm%7BJ%7D)