Question

Can you answer this Algebra question please?
(Just A, thank you).

Answer 1

`</p><p>d_f=d_i+v_0t+(1)/(2)at^2 \\</p><p>d_f-d_i-(1)/(2)at^2=v_0t \\</p><p>v_0=\boxed{(d_f+d_i-(1)/(2)at^2)/(t)} \\</p><p>`

Hope this helps.

Answer 2

A).
`v_(0)=(1)/(t)(d_(f)-d_(t)-(1)/(2)at^(2))`

B).
`t=\sqrt{(2)/(a)(d_(f)-d_(t))}`

Explanation:

The given expression is

`d_(f)=d_(t)+v_(0)t+(1)/(2)at^(2)`

A). We have to solve this expression for the value of
`v_(0)`

`d_(f)-d_(t)=v_(0)t+(1)/(2)at^(2)`

`d_(f)-d_(t)-(1)/(2)at^(2)=v_(0)t`

`v_(0)=(1)/(t)(d_(f)-d_(t)-(1)/(2)at^(2))`

B). If
`v_(0)=0` then we have to find the value of t.

We plug in the value of
`v_(0)=0` in the equation in Part A.

`0=(1)/(t)(d_(f)-d_(t)-(1)/(2)at^(2))`

`d_(f)-d_(t)-(1)/(2)at^(2)=0`

`(1)/(2)at^(2)=d_(f)-d_(t)`

`t^(2)=(2)/(a)(d_(f)-d_(t))`

`t=\sqrt{(2)/(a)(d_(f)-d_(t))}`