Answer:
The horse's final speed is 8.31 m/s
Explanation:
Motion With Constant Acceleration
It's a type of motion in which the velocity of an object changes uniformly in time.
The equation that describes the change of velocities is:
[tex]v_f=v_o+at\qquad\qquad [1][/tex]
Where:
a = acceleration
vo = initial speed
vf = final speed
t = time
The distance traveled by the object is calculated as follows:
[tex]\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2][/tex]
Solving [1] for t and substituting into [2] we get the following equation:
[tex]V_f^2=V_o^2+2aX[/tex]
Now we use the data provided by the question. The horse starts at vo=3 m/s and then accelerates at [tex]a=2 m/s^2[/tex] for a distance of x=15 m. The final speed is:
[tex]V_f^2=3^2+2(2)(15)[/tex]
[tex]V_f^2=9+60=69[/tex]
[tex]V_f=\sqrt{69}[/tex]
[tex]V_f=8.31\ m/s[/tex]
The horse's final speed is 8.31 m/s
