An example is free fall ,
Answer:
The average acceleration during the 6.0 s interval was -27 m/s².
Explanation:
Hi there!
The average acceleration is defined as the change in velocity over time:
a = Δv/t
Where:
a = acceleration.
Δv = change in velocity = final velocity - initial velocity
t = elapsed time
The change in velocity will be:
Δv = final velocity - initial velocity
Δv = -74 m/s - 87 m/s = -161 m/s
(notice the negative sign of the velocity that is in opposite direction to the direction considered positive)
Then the average acceleration will be:
a = Δv/t
a = -161 m/s / 6.0 s
a = -27 m/s²
The average acceleration during the 6.0 s interval was -27 m/s².
In voluntay smoth tissue muscles
<span>vf^2 = vi^2 + 2*a*d
---
vf = velocity final
vi = velocity initial
a = acceleration
d = distance
---
since the airplane is decelerating to zero, vf = 0
---
0 = 55*55 + 2*(-2.5)*d
d = (-55*55)/(2*(-2.5))
d = 605 meters
</span>
