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
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 ?
3 answers
v = dx/dt = 5.3 - 22 t
a = d^2x/dt^2 = -22 m/s^2
a = d^2x/dt^2 = -22 m/s^2
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!
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!
1 answer