Question

In the previous question, the person had an initial velocity of 8m/s and a constant acceleration of −4m/s2. How would the maximum distance he travels to the right of the origin change if instead his initial velocity were doubled (vx,0=16m/s)?

Answer 1

Answer:

Distance maximum

Xmax = 32m

Explanation:

Kinematics equation between velocities and distance:

$Vf^{2} =Vo^{2} +2ax$

For the distance maximum the final velocity is zero:

$x = Xmax$ ⇒ $Vf=0$

then, the equation becomes:

$Xmax= - \frac{Vo^{2}}{2a}$

We replace the values for the initial velocity and acceleration

Vo=16m/s; a = -4m/s2

Finally:

$Xmax=16^{2} /8=32m$