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x=v_(@)t+(1)/(2)at^(2)
The horizontal displacement, 
x, of an object with constant acceleration, 
a, initial velocity, 
v_(@), at elapsed time, 
t, is given by the equation. Which of the following equations correctly shows the acceleration in terms of displacement, initial velocity, and time?
Choose 1 answer:
(A) 
a=(2sqrt(x-v_(@)t))/(t)
(B) 
a=(2(x-v_(0)))/(t)
(C) 
a=(2(x-v_(@)t))/(t^(2))
(D) 
a=(2x)/(v_(0)t^(3))

x=vt+12at2 x=v_{\circ} t+\frac{1}{2} a t^{2} \newlineThe horizontal displacement, x x , of an object with constant acceleration, a a , initial velocity, v v_{\circ} , at elapsed time, t t , is given by the equation. Which of the following equations correctly shows the acceleration in terms of displacement, initial velocity, and time?\newlineChoose 11 answer:\newline(A) a=2xvtt a=\frac{2 \sqrt{x-v_{\circ} t}}{t} \newline(B) a=2(xv0)t a=\frac{2\left(x-v_{0}\right)}{t} \newline(C) a=2(xvt)t2 a=\frac{2\left(x-v_{\circ} t\right)}{t^{2}} \newline(D) a=2xv0t3 a=\frac{2 x}{v_{0} t^{3}}

Full solution

Q. x=vt+12at2 x=v_{\circ} t+\frac{1}{2} a t^{2} \newlineThe horizontal displacement, x x , of an object with constant acceleration, a a , initial velocity, v v_{\circ} , at elapsed time, t t , is given by the equation. Which of the following equations correctly shows the acceleration in terms of displacement, initial velocity, and time?\newlineChoose 11 answer:\newline(A) a=2xvtt a=\frac{2 \sqrt{x-v_{\circ} t}}{t} \newline(B) a=2(xv0)t a=\frac{2\left(x-v_{0}\right)}{t} \newline(C) a=2(xvt)t2 a=\frac{2\left(x-v_{\circ} t\right)}{t^{2}} \newline(D) a=2xv0t3 a=\frac{2 x}{v_{0} t^{3}}
  1. Given equation for displacement: We start with the given equation for horizontal displacement: x=v@t+12at2x = v_{@}t + \frac{1}{2}at^2.
  2. Isolate acceleration: To solve for acceleration, aa, we need to isolate it on one side of the equation. First, subtract v(@)tv_(@)t from both sides: xv(@)t=12at2x - v_(@)t = \frac{1}{2}at^2.
  3. Multiply by 22: Next, multiply both sides by 22 to get rid of the fraction: 2(xv@t)=at22(x - v_{@}t) = at^2.
  4. Divide by t2t^2: Now, divide both sides by t2t^2 to solve for aa: a=2(xv@t)t2a = \frac{2(x - v_{@}t)}{t^2}.

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