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
Noah G
May 22, 2018

The value of #v# is #11/2#

Explanation:

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

#int 4 +x dx = 1/2x^2 + 4x + C#

We can now solve for #C#.

#1/2(0)^2 + 4(0) + C =1#

#C = 1#

Therefore the velocity function is #v= 1/2x^2 + 4x+ 1# and the value of #v# at #x = 1# is hence #v(1) = 1/2 + 4 + 1 = 11/2#

Hopefully this helps!

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