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

The horse's final speed is 8.31 m/s

Step-by-step 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+(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=√(69)`

`V_f=8.31\ m/s`

The horse's final speed is 8.31 m/s