Answer to Question #177683 in Math for John

Question

If s = 2t³ − 21t² +60t describes the displacement of a body at time t from a fixed point O on a straight line, find:

(A). s when t = 3 seconds

b)         the values of t when the body is at rest

c)         the initial velocity of the body

d)         an expression for the acceleration of the body at time t

e)         the initial acceleration of the body.   

Answer 1

$s(t)=2t^3-21t^2+60t$

a) $s(3)=2\cdot 3^3-21\cdot 3^2+60\cdot 3=54-189+180=45$

b) The body is at rest when $v(t)=0$ .

$v(t)=s^\prime (t)=6t^2-42t+60=0$

$6t^2-42t+60=0$

$t^2-7t+10=0$

$D=(-7)^2-4\cdot10=9$

$t_{1,2}=\frac{7\pm \sqrt{9}}{2}$

$t_1=2$ and $t_2=5$

c) $v(0)=6\cdot 0^2-42\cdot 0+60=60$

d) $a(t)=s^{\prime \prime}(t)=v^\prime (t)=12t-42$

e) $a(0)=12\cdot 0-42=-42$

Answer: a) $s(3)=45$ ; b) $t_1=2$ and $t_2=5$ ; c) $v(0)=60$ ; d) $a(t)=12t-42$ ; e) $a(0)=-42$ .