In a displacement/time graph, the slope of the line is equal to the velocity
To solve this problem it is necessary to apply the kinematic equations of motion.
By definition we know that the position of a body is given by
![x=x_0+v_0t+at^2](https://tex.z-dn.net/?f=x%3Dx_0%2Bv_0t%2Bat%5E2)
Where
Initial position
Initial velocity
a = Acceleration
t= time
And the velocity can be expressed as,
![v_f = v_0 + at](https://tex.z-dn.net/?f=v_f%20%3D%20v_0%20%2B%20at)
Where,
![v_f = Final velocity](https://tex.z-dn.net/?f=v_f%20%3D%20Final%20velocity)
For our case we have that there is neither initial position nor initial velocity, then
![x= at^2](https://tex.z-dn.net/?f=x%3D%20at%5E2)
With our values we have
, rearranging to find a,
![a=\frac{x}{t^2}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bx%7D%7Bt%5E2%7D)
![a = \frac{ 401.4}{4.945^2}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B%20401.4%7D%7B4.945%5E2%7D)
![a = 16.41m/s^2](https://tex.z-dn.net/?f=a%20%3D%2016.41m%2Fs%5E2)
Therefore the final velocity would be
![v_f = v_0 + at](https://tex.z-dn.net/?f=v_f%20%3D%20v_0%20%2B%20at)
![v_f = 0 + (16.41)(4.945)](https://tex.z-dn.net/?f=v_f%20%3D%200%20%2B%20%2816.41%29%284.945%29)
![v_f = 81.14m/s](https://tex.z-dn.net/?f=v_f%20%3D%2081.14m%2Fs)
Therefore the final velocity is 81.14m/s
Newton's 2nd law of motion:
Force = (mass) x (acceleration)
= (0.314 kg) x (164 m/s²)
= 51.5 newtons
(about 11.6 pounds).
Notice that the ball is only accelerating while it's in contact with the racket. The instant the ball loses contact with the racket, it stops accelerating, and sails off in a straight line at whatever speed it had when it left the strings.
~ I hope this helped, and I would appreciate Brainliest. ♡ ~ ( I request this to all the lengthy answers I give ! )
Answer:
Fa = 5000 [N]
Explanation:
To solve this problem we must use Newton's second law, which tells us that the sum of forces on a body is equal to the product of mass by acceleration.
Let's assume that the movement of the plane is to the right, any movement or force to the right will be marked with a positive sign, while any force or movement to the left, will be taken as negative.
The force of the turbine drives the plane to the right, therefore it is positive, the acceleration is constant and keeps the movement to the right, therefore it is positive, the wind drag force tries to prevent the movement of the plane to the left therefore it is negative, with this analysis we deduce the following equation.
ΣF = m*a
where:
ΣF = sum of forces [N] (units of Newtons)
m = mass = 65000 [kg]
a = acceleration = 3 [m/s²]
Fa = force exerted by the air [N]
200000 - Fa = 65000*3
Fa = 200000 - (3*65000)
Fa = 5000 [N]