x=v∘t+21at2The 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=t2x−v∘t(B) a=t2(x−v0)(C) a=t22(x−v∘t)(D) a=v0t32x
Q. x=v∘t+21at2The 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=t2x−v∘t(B) a=t2(x−v0)(C) a=t22(x−v∘t)(D) a=v0t32x
Given equation for displacement: We start with the given equation for horizontal displacement: x=v@t+21at2.
Isolate acceleration: To solve for acceleration, a, we need to isolate it on one side of the equation. First, subtract v(@)t from both sides: x−v(@)t=21at2.
Multiply by 2: Next, multiply both sides by 2 to get rid of the fraction: 2(x−v@t)=at2.
Divide by t2: Now, divide both sides by t2 to solve for a: a=t22(x−v@t).
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