Question

Bill steps off a 3.0-m-high diving board and drops to the water below. At the same time, Ted jumps upward with a speed of 4.2 m/s from a 1.0-m-high diving board. Choosing the origin to be at the water's surface, and upward to be the positive x direction, write x-versus-t equations of motion for both Bill and Ted.

Answer 1

Answer:

equation of  motion for Bill is

$y(t) = 4.9t^2$

equation of  motion for Ted is

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

Explanation:

Taking downward position positive and upward position negative

g = 9.8 m/s^2

equation of  motion for Bill is

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 0 + 0(t) +\frac{1}{2}gt^2$

$y(t) = \frac{1}{2}\times (9.8t)^2$

$y(t) = 4.9t^2$

equation of  motion for Ted is

$y_0 = 2m -1m = 2m$

$y_0 = -4.2 m/s$

$y(t) = y_0 +v_0 t +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}gt^2$

$y(t) = 2 + (-4.2)(t) +\frac{1}{2}\times (9.8t)^2$

$y(t) = 2 + (-4.2)(t) + 4.9t^2$

Answer 2

Answer:

Answer:

For Bill:

$x(t)=3-(4.9*t^{2})$

For Ted:

$x(t)=1+(4.2*t)+(-4.9*t^{2} )$

Explanation:

For Bill:

$Initial position=x_{0}=3$

$Initial velocity=v_{0}=0$

Now using $2^{nd}$ equation of motion,we have

$x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})$

$x_{0} =3$  ,$v_{0}=0$

Thus,equation becomes

$x-3=1/2*g*t^{2}$

$x=3+(0.5*g*t^{2})$

Taking acceleration upward positive and downward negative.

$g=-10$ $m/s^{2}$

$x(t)=3-4.9*t^{2}$    for bill

For Ted

$x_{0} =1$

$v_{0}=4.2$ $m/s$

Using the same equation

$x-x_{0}=(v_{0} *t)+(1/2*g*t^{2})$

$x_{0}=1$ $m$

$v_{0}=4.2$ $m/s$

Substitute values

$x-1=(4.2*t)+(1/2*g*t^{2})$

$g=-10$ $m/s^{2}$

Thus equation becomes

$x(t)=1+(4.2*t)+(-4.9*t^{2})$   for Ted