Answer:
In a straight line
Explanation:
Why, because of the power of inerta
Answer:
34 m/s
Explanation:
m = Mass of glider with person = 680 kg
v = Velocity of glider with person = 34 m/s
= Mass of glider without person = 680-60 kg
= Gliders speed just after the skydiver lets go
= Mass of person = 60 kg
= Velcotiy of person = 34 m/s
As the linear momentum of the system is conserved
![m_1v_1+m_2v_2=mv\\\Rightarrow v_1=\dfrac{mv-m_2v_2}{m_1}\\\Rightarrow v_1=\dfrac{680\times 34-60\times 34}{680-60}\\\Rightarrow v_1=34\ m/s](https://tex.z-dn.net/?f=m_1v_1%2Bm_2v_2%3Dmv%5C%5C%5CRightarrow%20v_1%3D%5Cdfrac%7Bmv-m_2v_2%7D%7Bm_1%7D%5C%5C%5CRightarrow%20v_1%3D%5Cdfrac%7B680%5Ctimes%2034-60%5Ctimes%2034%7D%7B680-60%7D%5C%5C%5CRightarrow%20v_1%3D34%5C%20m%2Fs)
The gliders speed just after the skydiver lets go is 34 m/s
Answer:
The taken is ![t_A = 19.0 \ s](https://tex.z-dn.net/?f=t_A%20%20%3D%2019.0%20%5C%20s)
Explanation:
Frm the question we are told that
The speed of car A is ![v_A = 22 \ m/s](https://tex.z-dn.net/?f=v_A%20%20%3D%20%2022%20%5C%20m%2Fs)
The speed of car B is ![v_B = 29.0 \ m/s](https://tex.z-dn.net/?f=v_B%20%20%3D%2029.0%20%5C%20m%2Fs)
The distance of car B from A is ![d = 300 \ m](https://tex.z-dn.net/?f=d%20%3D%20300%20%5C%20m)
The acceleration of car A is ![a_A = 2.40 \ m/s^2](https://tex.z-dn.net/?f=a_A%20%20%3D%202.40%20%5C%20m%2Fs%5E2)
For A to overtake B
The distance traveled by car B = The distance traveled by car A - 300m
Now the this distance traveled by car B before it is overtaken by A is
![d = v_B * t_A](https://tex.z-dn.net/?f=d%20%3D%20v_B%20%2A%20t_A)
Where
is the time taken by car B
Now this can also be represented as using equation of motion as
![d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300](https://tex.z-dn.net/?f=d%20%3D%20v_A%20t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7Da_A%20t_A%5E2%20-%20300)
Now substituting values
![d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=d%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
Equating the both d
![v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=v_B%20%2A%20t_A%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
substituting values
![29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=29%20%2A%20t_A%20%3D%2022t_A%20%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
![7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300](https://tex.z-dn.net/?f=7%20t_A%20%3D%20%5Cfrac%7B1%7D%7B2%7D%20%282.40%29%5E2%20t_A%5E2%20-%20300)
![7 t_A =1.2 t_A^2 - 300](https://tex.z-dn.net/?f=7%20t_A%20%3D1.2%20t_A%5E2%20-%20300)
![1.2 t_A^2 - 7 t_A - 300 = 0](https://tex.z-dn.net/?f=1.2%20t_A%5E2%20-%207%20t_A%20-%20300%20%20%3D%200)
Solving this using quadratic formula we have that
![t_A = 19.0 \ s](https://tex.z-dn.net/?f=t_A%20%20%3D%2019.0%20%5C%20s)
<span>A. crest, crest
hope im right!(: </span>