a horse runs with an initial velocity of 11m/s and slows to 5.2 m/s over a time interval of 3.1 s what is the horse's average acceleration
how much time is required for an airplane to reach its takeoff speed of 77m/s it is starts from rest and its average acceleration is 8.2 m/s
a = change in v / change in time
= (5.2 - 11) / 3.1
= -1.87 m/s^2
fgy
Change in time = change in v / average acceleration
change in t = (77m/s - 0 m/s) / 8.2 m/s^2
=9.39 s or 9.4 s
Well, it seems like this horse needs a bit of training in the speed department! To find the horse's average acceleration, we can use the formula:
acceleration = (change in velocity) / (change in time)
The change in velocity is the difference between the final velocity (5.2 m/s) and the initial velocity (11 m/s), which is -5.8 m/s (negative because the horse is slowing down). The change in time is 3.1 seconds.
So, the horse's average acceleration would be -5.8 m/s divided by 3.1 s. But I gotta say, this horse is going backward when it comes to acceleration!
To find the horse's average acceleration, we can use the equation:
average acceleration = (final velocity - initial velocity) / time interval
Given:
Initial velocity (u) = 11 m/s
Final velocity (v) = 5.2 m/s
Time interval (t) = 3.1 s
Substituting the given values into the equation, we get:
average acceleration = (5.2 m/s - 11 m/s) / 3.1 s
Now, we can calculate the average acceleration:
average acceleration = (-5.8 m/s) / 3.1 s
Therefore, the horse's average acceleration is approximately -1.87 m/s².
Note that the negative sign indicates that the horse is decelerating or slowing down.