(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?

Question

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.

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Answer 1 · 2018-04-26T14:49:19

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!