Question

A horse leaves its stable for a morning run. It begins its morning trot at 3.25 m/sec trot, and soon breaks into a quick 6.4 m/sec gallop, taking only 2.31 sec to reach this speed. How far did the horse travel during this period of acceleration?

Answer 1

The horse traveled approximately 12.01 meters during its period of acceleration from a trot at 3.25 m/s to a gallop at 6.4 m/s over 2.31 seconds.

Step-by-step explanation:

Calculating Distance During AccelerationTo find out how far the horse traveled during its period of acceleration, we can use the formula for the distance covered under constant acceleration, which is d = ut + (1/2)at2, where u is the initial velocity, a is the acceleration, and t is the time.

First, we need to calculate the acceleration using the initial and final velocities and the time taken. We have: initial velocity (u) = 3.25 m/sec, final velocity (v) = 6.4 m/sec, and time (t) = 2.31 sec. The acceleration (a) can be found using the formula a = (v - u) / t. Plugging in the values, we get:

a = (6.4 m/s - 3.25 m/s) / 2.31 s = 1.36 m/s2

Now, we use the calculated acceleration to find the distance using the distance formula mentioned above:

d = (3.25 m/s * 2.31 s) + (1/2 * 1.36 m/s2 * (2.31 s)2)

After calculating, we find that the distance d is approximately 12.01 meters.