The position of a ball as a function of time is given by x=(5.3m/s)t + (−11m/s^2)t^2
What would be the acceleration of the ball, and the velocity from t=0 to t=1s ?
Given that the position is x=(5.3m/s)t + (−11m/s^2)t^2
The acceleration can be found by adding the two numbers (5.3 - 11) to get -5.7 m/s as the velocity.
Next, you can get the acceleration by using this formula: v = dx/dt
which gives you -5.7 - 22 t
rearrange the formula to get 3.8 m/s ^2 as acceleration!
Well, well, well, we've got ourselves a ball and some time! Let's see what we can do here.
To find the acceleration, we take the second derivative of the position function. So, grab your calculators and get ready for some number crunching!
The first derivative of x with respect to time (t) gives us the velocity (v):
v = d/dt[(5.3m/s)t + (-11m/s^2)t^2]
v = 5.3m/s - 22m/s^2t
Now, to find the acceleration (a), we take the derivative of the velocity function with respect to time:
a = d/dt[5.3m/s - 22m/s^2t]
a = -22m/s^2
So, the acceleration of the ball is a constant -22m/s^2. That's quite a zippy little ball!
Now, let's find the velocity of our little ball from t=0 to t=1s. Plug in the values and watch the magic happen!
v(1s) = 5.3m/s - 22m/s^2(1s)
v(1s) = 5.3m/s - 22m/s^2
v(1s) = -(16.7m/s)
So, the velocity of the ball at t=1s is -16.7m/s. Negative velocity? Well, I guess our ball is feeling a bit down in the dumps today.
Hope that helps! Now go forth and conquer the world of physics, my friend!
To determine the acceleration of the ball, we need to find the second derivative of the position function with respect to time (t).
Given the position function, x = (5.3m/s)t - (11m/s^2)t^2
1. Calculate the first derivative of x with respect to t to find the velocity function, v(t).
v(t) = d(x)/dt = (d/dt)(5.3t - 11t^2)
= 5.3 - 22t
2. Calculate the second derivative of x with respect to t to find the acceleration function, a(t).
a(t) = d^2(x)/dt^2 = (d/dt)(5.3 - 22t)
= -22
So, the acceleration of the ball is constant and equal to -22 m/s^2.
To find the velocity from t = 0 to t = 1s, we can substitute these values into the velocity function, v(t).
At t = 0s: v(0) = 5.3 - 22(0) = 5.3 m/s
At t = 1s: v(1) = 5.3 - 22(1) = -16.7 m/s
Therefore, the velocity of the ball at t = 0s is 5.3 m/s, and the velocity at t = 1s is -16.7 m/s.
the acceleration is ... -11 m/s^2
the average velocity from t=0 to t=1s
... [5.3 m/s + (5.3 m/s - 11 m/s)] / 2
v = dx/dt = 5.3 - 22 t
a = d^2x/dt^2 = -22 m/s^2