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:

For Bill:

`x(t)=3-(4.9*t^(2))`

For Ted:

`x(t)=1+(4.2*t)+(-4.9*t^(2) )`

Step-by-step 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

Answer 2

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`

Step-by-step 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 +(1)/(2)gt^2`

`y(t) = 0 + 0(t) +(1)/(2)gt^2`

`y(t) = (1)/(2)* (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 +(1)/(2)gt^2`

`y(t) = 2 + (-4.2)(t) +(1)/(2)gt^2`

`y(t) = 2 + (-4.2)(t) +(1)/(2)* (9.8t)^2`

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