A horse is running across a field. The horse accelerates at 8.00 m/s2 over a distance of 9.70 m. It runs until its speed is 23.4 m/s and continues at this speed. What was the horse's initial speed?
v = Vi + a t
23.4 = Vi + 8 t
average speed = (Vi+23.4)/2
so
t = 9.7/[ (Vi+23.4)/2 ]
so
23.4 = Vi + 8*9.7/[ (Vi+23.4)/2 ]
solve for Vi
To find the horse's initial speed, we can use the equation of motion:
v^2 = u^2 + 2as
where:
- v is the final velocity (23.4 m/s),
- u is the initial velocity (what we want to find),
- a is the acceleration (8.00 m/s^2), and
- s is the distance covered during acceleration (9.70 m).
We can rearrange the equation to solve for u:
u^2 = v^2 - 2as
Now, let's substitute the given values into the equation:
u^2 = (23.4 m/s)^2 - 2 * (8.00 m/s^2) * (9.70 m)
u^2 = 547.56 m^2/s^2 - 155.2 m^2/s^2
u^2 = 392.36 m^2/s^2
To find u, we take the square root of both sides:
u = √392.36 m^2/s^2
u ≈ 19.8 m/s
Therefore, the horse's initial speed was approximately 19.8 m/s.