Question

. A particle moves a fixed point O. The velocity v of the particle at time t s is given by v = (3t − 2)(t − 4) and s . A particle moves a fixed point O. The velocity v of the particle at time t s is given by

v = (3t − 2)(t − 4) and s = 8m when t = 1s, find:

(a) the initial velocity of the particle [1]

(b) the acceleration of the particle when t = 3s. [2]

(c) the values of t when the particle is at rest [2]

(d) the distance the particle is from O when t = 2s. [5]

(e) the distance travelled when t = 2s= 8m when t = 1s, find: (a) the initial velocity of the particle [1] (b) the acceleration of the particle when t = 3s. [2] (c) the values of t when the particle is at rest [2] (d) the distance the particle is from O when t = 2s. [5] (e) the distance travelled when t = 2s

Answer 1

$v=(3t-2)(t-4)=3t^2-14t+8,$

$s(t)=\int vdt=t^3-7t^2+8t+C,$

$s(1)=1-7+8+C=2+C=8,\implies C=6,$

a)

$s(0)=6~m,$

b)

$a(t)=v'(t)=6t-14,$

$a(3)=18-14=4~\frac{m}{s^2},$

c)

$v=0: t=\frac 23~s~\text{or}~t=4~s,$

d)

$s(2)=8-28+16+6=2~m,$

e)

$s(1)=1-7+8+6=8~m.$