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If the velocity of a particle moving along the
x-axis is
v(t)=2t-4 and if at
t=0 its position is 4 , then at any time
t its position
x(t) is
(A)
t^(2)-4t
(B)
t^(2)-4t-4
(C)
t^(2)-4t+4
(D)
2t^(2)-4t
(E)
2t^(2)-4t+4

If the velocity of a particle moving along the $x$ -axis is $v(t)=2 t-4$ and if at $t=0$ its position is $4$ , then at any time $t$ its position $x(t)$ is $\newline$ (A) $t^{2}-4 t$ $\newline$ (B) $t^{2}-4 t-4$ $\newline$ (C) $t^{2}-4 t+4$ $\newline$ (D) $2 t^{2}-4 t$ $\newline$ (E) $2 t^{2}-4 t+4$

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Relate position, velocity, speed, and acceleration using derivatives

Full solution

Q. If the velocity of a particle moving along the

x

-axis is

v(t)=2 t-4

and if at

t=0

its position is

4

, then at any time

t

its position

x(t)

is

\newline

(A)

t^{2}-4 t

\newline

(B)

t^{2}-4 t-4

\newline

(C)

t^{2}-4 t+4

\newline

(D)

2 t^{2}-4 t

\newline

(E)

2 t^{2}-4 t+4

Integrate velocity function: First, we need to integrate the velocity function $v(t) = 2t - 4$ to find the position function $x(t)$ .
Find position function: The integral of $2t$ is $t^2$ , and the integral of $-4$ is $-4t$ . So, the integral of $v(t) = 2t - 4$ is $x(t) = t^2 - 4t + C$ , where $C$ is the constant of integration.
Determine constant of integration: We know that at $t=0$ , the position $x(0)$ is $4$ . So, we plug in $t=0$ into $x(t)$ to find $C$ : $4 = (0)^2 - 4(0) + C$ .
Calculate position at $t=0$ : Solving for $C$ , we get $C = 4$ .
Finalize position function: Now we have the position function $x(t) = t^2 - 4t + 4$ .
Verify matching option: We check the options given and find that option (C) matches our position function $x(t) = t^2 - 4t + 4$ .

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