Question

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

Answer 1

$d_f=d_i+v_0t+\dfrac{1}{2}at^2 \\d_f-d_i-\dfrac{1}{2}at^2=v_0t \\v_0=\boxed{\dfrac{d_f+d_i-\frac{1}{2}at^2}{t}}$

Hope this helps.

Answer 2

Answer:

A). $v_{0}=\frac{1}{t}(d_{f}-d_{t}-\frac{1}{2}at^{2})$

B). $t=\sqrt{\frac{2}{a}(d_{f}-d_{t})}$

Step-by-step explanation:

The given expression is

$d_{f}=d_{t}+v_{0}t+\frac{1}{2}at^{2}$

A). We have to solve this expression for the value of $v_{0}$

$d_{f}-d_{t}=v_{0}t+\frac{1}{2}at^{2}$

$d_{f}-d_{t}-\frac{1}{2}at^{2}=v_{0}t$

$v_{0}=\frac{1}{t}(d_{f}-d_{t}-\frac{1}{2}at^{2})$

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.

$0=\frac{1}{t}(d_{f}-d_{t}-\frac{1}{2}at^{2})$

$d_{f}-d_{t}-\frac{1}{2}at^{2}=0$

$\frac{1}{2}at^{2}=d_{f}-d_{t}$

$t^{2}=\frac{2}{a}(d_{f}-d_{t})$

$t=\sqrt{\frac{2}{a}(d_{f}-d_{t})}$