Question

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?

Answer 1

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!