Answer:
positive
Explanation:
The ball is rolling down with a negative velocity, but the velocity is slowing down. therefore the velocity must increase in order for the ball to slow down.
For example let the ball's initial velocity be -15 m/s. and it is slowing down to let's say -13 m/s. Well this means that it's velocity has increase by 2 m/s. So, its acceleration is positive.
Answer:
y = -19.2 sin (23.15t) cm
Explanation:
The spring mass system is an oscillatory movement that is described by the equation
y = yo cos (wt + φ)
Let's look for the terms of this equation the amplitude I
y₀ = 19.2 cm
Angular velocity is
w = √ (k / m)
w = √ (245 / 0.457
w = 23.15 rad / s
The φ phase is determined for the initial condition t = 0 s
, the velocity is negative v (0) = -vo
The speed of the equation is obtained by the derivative with respect to time
v = dy / dt
v = - y₀ w sin (wt + φ)
For t = 0
-vo = -yo w sin φ
The angular and linear velocity are related v = w r
v₀ = w r₀
v₀ = v₀ sinφ
sinφ = 1
φ = sin⁻¹ 1
φ = π / 4 rad
Let's build the equation
y = 19.2 cos (23.15 t + π/ 4)
Let's use the trigonometric ratio π/ 4 = 90º
Cos (a +90) = cos a cos90 - sin a sin sin 90 = 0 - sin a
y = -19.2 sin (23.15t) cm
force is mass * acceleration
so 2kg * 2m/s^2 = 4 N
Let
denote the position vector of the ball hit by player A. Then this vector has components
![\begin{cases}r_{Ax}=\left(2.4\,\frac{\mathrm m}{\mathrm s}\right)t\\r_{Ay}=1.2\,\mathrm m-\frac12gt^2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dr_%7BAx%7D%3D%5Cleft%282.4%5C%2C%5Cfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29t%5C%5Cr_%7BAy%7D%3D1.2%5C%2C%5Cmathrm%20m-%5Cfrac12gt%5E2%5Cend%7Bcases%7D)
where
is the magnitude of the acceleration due to gravity. Use the vertical component
to find the time at which ball A reaches the ground:
![1.2\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.49\,\mathrm s](https://tex.z-dn.net/?f=1.2%5C%2C%5Cmathrm%20m-%5Cdfrac12%5Cleft%289.8%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%3D0%5Cimplies%20t%3D0.49%5C%2C%5Cmathrm%20s)
The horizontal position of the ball after 0.49 seconds is
![\left(2.4\,\dfrac{\mathrm m}{\mathrm s}\right)(0.49\,\mathrm s)=12\,\mathrm m](https://tex.z-dn.net/?f=%5Cleft%282.4%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D%5Cright%29%280.49%5C%2C%5Cmathrm%20s%29%3D12%5C%2C%5Cmathrm%20m)
So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector
of the ball hit by player B has
![\begin{cases}r_{Bx}=v_0t\\r_{By}=1.6\,\mathrm m-\frac12gt^2\end{cases}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7Dr_%7BBx%7D%3Dv_0t%5C%5Cr_%7BBy%7D%3D1.6%5C%2C%5Cmathrm%20m-%5Cfrac12gt%5E2%5Cend%7Bcases%7D)
Again, we solve for the time it takes the ball to reach the ground:
![1.6\,\mathrm m-\dfrac12\left(9.8\,\dfrac{\mathrm m}{\mathrm s^2}\right)t^2=0\implies t=0.57\,\mathrm s](https://tex.z-dn.net/?f=1.6%5C%2C%5Cmathrm%20m-%5Cdfrac12%5Cleft%289.8%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%5E2%7D%5Cright%29t%5E2%3D0%5Cimplies%20t%3D0.57%5C%2C%5Cmathrm%20s)
After this time, we expect a horizontal displacement of 12 meters, so that
satisfies
![v_0(0.57\,\mathrm s)=12\,\mathrm m](https://tex.z-dn.net/?f=v_0%280.57%5C%2C%5Cmathrm%20s%29%3D12%5C%2C%5Cmathrm%20m)
![\implies v_0=21\,\dfrac{\mathrm m}{\mathrm s}](https://tex.z-dn.net/?f=%5Cimplies%20v_0%3D21%5C%2C%5Cdfrac%7B%5Cmathrm%20m%7D%7B%5Cmathrm%20s%7D)
Answer:
The value is ![a_t = 2.42 \ m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%20%202.42%20%5C%20%20m%2Fs%5E2)
Explanation:
From the question we are told that
The radius of the tires is ![r = 0.22 \ m](https://tex.z-dn.net/?f=r%20%3D%20%200.22%20%5C%20%20m)
The angular acceleration is ![\alpha = 11.0 \ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%20%3D%20%2011.0%20%5C%20rad%2Fs%5E2)
Generally the linear acceleration is mathematically represented as
![a_t = r * \alpha](https://tex.z-dn.net/?f=a_t%20%3D%20%20r%20%2A%20%20%5Calpha)
=> ![a_t = 0.22 * 11](https://tex.z-dn.net/?f=a_t%20%3D%20%200.22%20%20%2A%20%2011)
=> ![a_t = 2.42 \ m/s^2](https://tex.z-dn.net/?f=a_t%20%3D%20%202.42%20%5C%20%20m%2Fs%5E2)