Answer:
A
Explanation:
ignore this on...............................
To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.
From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by
![v_f^2=v_i^2+2ax](https://tex.z-dn.net/?f=v_f%5E2%3Dv_i%5E2%2B2ax)
Where,
Final velocity
Initial Velocity
a = Acceleration
x = Displacement
Acceleration can be expressed in terms of the drag coefficient by means of
Frictional Force
Force by Newton's second Law
Where,
m = mass
a= acceleration
Kinetic frictional coefficient
g = Gravity
Equating both equation we have that
![F_f = F](https://tex.z-dn.net/?f=F_f%20%3D%20F)
![\mu_k mg=ma](https://tex.z-dn.net/?f=%5Cmu_k%20mg%3Dma)
![a = \mu_k g](https://tex.z-dn.net/?f=a%20%3D%20%5Cmu_k%20g)
Therefore,
![v_f^2=v_i^2+2ax](https://tex.z-dn.net/?f=v_f%5E2%3Dv_i%5E2%2B2ax)
![0=v_i^2+2(\mu_k g)x](https://tex.z-dn.net/?f=0%3Dv_i%5E2%2B2%28%5Cmu_k%20g%29x)
Re-arrange to find x,
![x = \frac{v_i^2}{2(-\mu_k g)}](https://tex.z-dn.net/?f=x%20%3D%20%5Cfrac%7Bv_i%5E2%7D%7B2%28-%5Cmu_k%20g%29%7D)
The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.
Answer:
Angle: ![48.19^o](https://tex.z-dn.net/?f=48.19%5Eo)
Explanation:
<u>Two-Dimension Motion</u>
When the object is moving in one plane, the velocity, acceleration, and displacement are vectors. Apart from the magnitudes, we also need to find the direction, often expressed as an angle respect to some reference.
Our boy can swim at 3 m/s from west to east in still water and the river he's attempting to cross interacts with him at 2 m/s southwards. The boy will move east and south and will reach the other shore at a certain distance to the south from where he started. It happens because there is a vertical component of his velocity that is not compensated.
To compensate for the vertical component of the boy's speed, he only has to swim at a certain angle east of the north (respect to the shoreline). The goal is to make the boy's y component of his velocity equal to the velocity of the river. The vertical component of the boy's velocity is
![v_b\ cos\alpha](https://tex.z-dn.net/?f=v_b%5C%20cos%5Calpha)
where
is the speed of the boy in still water and
is the angle respect to the shoreline. If the river flows at speed
, we now set
![v_b\ cos\alpha=v_s](https://tex.z-dn.net/?f=v_b%5C%20cos%5Calpha%3Dv_s)
![\displaystyle cos\alpha=\frac{v_s}{v_b}=\frac{2}{3}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20cos%5Calpha%3D%5Cfrac%7Bv_s%7D%7Bv_b%7D%3D%5Cfrac%7B2%7D%7B3%7D)
![\alpha=48.19^o](https://tex.z-dn.net/?f=%5Calpha%3D48.19%5Eo)