Question

(i) Show that the value of **k** is 0.1. (ii) Find an expression for the displacement of the particle from the origin in terms of t?

A particle starts from a fixed origin with a velocity `0.4ms^-1` and moves in a straight line. The acceleration a `ms^-1` of the particle t s after it leaves the origin is given by `a = k(3t^2 - 12t + 2)`, where k is a constant. when t =1, the velocity of P is `0.1ms^-1.`
(i) Show that the value of k is 0.1.
(ii) Find an expression for the displacement of the particle from the origin in terms of t.

Answer 1

Recall that the velocity of a particle is the integral of the accerlation function.

`v = k(t^3 - 6t^2 + 2t )+ C`

Since `v(0) = 0.4`, we can see that `C = 0.4`

`v = k(t^3 - 6t^2 + 2t) + 0.4`

The question states that `v(1) = 0.1`.

`0.1 = k(1^3 - 6(1)^2 + 2(1)) + 0.4`

`0.1 = k(-3) + 0.4`

`-0.3 = -3k`

`k = 0.1`

As required.

b) We must take the anti-derivative of the velocity function to get the position function.

`d = 0.1(1/4t^4 - 2t^3 + t^2) + C`

We know that `C = 0` because the particle begins at the origin. Thus

`d = 1/10(1/4t^4 - 2t^3 + t^2)`

`d = 1/40t^4 - 1/5t^3 + 1/10t^2`

Hopefully this helps!