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)
Answer:
0.0665 days
Explanation:
We are given;
The mean distance from the Earth's center to the moon;a1 = 385000 km
The mean distance from the Earth's center to the space craft;a2 = 6965 km
Formula for kepplers third law is;
T² = 4π²a³/GM
However, the proportion of both distances would be;
(T1)²/(T2)² = (a1)³/(a2)³
Where;
T1 is the period of orbit of the moon around the earth. T1 has a standard value of 27.322 days
T2 is the period of the space craft orbit.
Making T2 the subject, we have;
T2 = √((T1)²×(a2)³)/(a1)³)
Thus, plugging in the relevant values;
T2 = √(27.322² × 6965³)/(385000)³
T2 = 0.0665 days
Answer:
, the kinetic energy of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.
Answer:
77.8s
Explanation:
Let d distance between the asteroid and space tug
So;d=Xtug+Xspace
Xtug=VtT+0.5atT^2
Xspace=VsT+0.5asT^2
Since Vt=Vs=0 initial velocity
Then
d=0.5(atT^2+asT^2)
T^2( at+as)=2d
T=√(2d/at+as)
But force F = mass M*acceleration a
Hence at=Ft/mt ,as=Fs/ms
But note Ft=F=Fs since the Same force acts on it
Hence T=√( 2d/F(1/mt+1/Ms))
T=√(2*493/366(1/3460+1/6430)
T=√(986/0.1627)=√(6060.195)=77.8s