A particle moves in a straight line so that at time t its displacement from a fixed origin is x and its velocity is v. If its acceleration is 4 + x and v = 1 when x = 0, what is v when x = 1?

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A particle moves in a straight line so that at time t its displacement from a fixed origin is x and its velocity is v. If its acceleration is 4 + x and v = 1 when x = 0, what is v when x = 1?

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Answer 1 · 2018-05-22T15:37:34

The value of $v$ $v$ is $\frac{11}{2}$ $11/2$

Explanation:

Recall that the derivative of velocity is acceleration and the anti-derivative of acceleration is velocity.

$\int 4 + x d x = \frac{1}{2} x^{2} + 4 x + C$ $int 4 +x dx = 1/2x^2 + 4x + C$

We can now solve for $C$ $C$ .

$\frac{1}{2} {(0)}^{2} + 4 (0) + C = 1$ $1/2(0)^2 + 4(0) + C =1$

$C = 1$ $C = 1$

Therefore the velocity function is $v = \frac{1}{2} x^{2} + 4 x + 1$ $v= 1/2x^2 + 4x+ 1$ and the value of $v$ $v$ at $x = 1$ $x = 1$ is hence $v (1) = \frac{1}{2} + 4 + 1 = \frac{11}{2}$ $v(1) = 1/2 + 4 + 1 = 11/2$

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A particle moves in a straight line so that at time t its displacement from a fixed origin is x and its velocity is v. If its acceleration is 4 + x and v = 1 when x = 0, what is v when x = 1?

1 Answer

Explanation:

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