Question

A horse trotting at 3 meters per second speeds up with an acceleration of 2
meters per second2 for a distance of 15 meters. Calculate the horse’s final
speed.

Answer 1

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:

$v_f=v_o+at\qquad\qquad [1]$

Where:

a   = acceleration

vo = initial speed

vf  = final speed

t    = time

The distance traveled by the object is calculated as follows:

$\displaystyle x=v_o.t+\frac{a.t^2}{2}\qquad\qquad [2]$

Solving [1] for t and substituting into [2] we get the following equation:

$V_f^2=V_o^2+2aX$

Now we use the data provided by the question. The horse starts at vo=3 m/s and then accelerates at $a=2 m/s^2$ for a distance of x=15 m. The final speed is:

$V_f^2=3^2+2(2)(15)$

$V_f^2=9+60=69$

$V_f=\sqrt{69}$

$V_f=8.31\ m/s$

The horse's final speed is 8.31 m/s