Question #177683

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.   


1
Expert's answer
2021-04-15T07:40:08-0400

s(t)=2t321t2+60ts(t)=2t^3-21t^2+60t


a) s(3)=2332132+603=54189+180=45s(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)=0v(t)=0 .

v(t)=s(t)=6t242t+60=0v(t)=s^\prime (t)=6t^2-42t+60=0

6t242t+60=06t^2-42t+60=0

t27t+10=0t^2-7t+10=0

D=(7)2410=9D=(-7)^2-4\cdot10=9

t1,2=7±92t_{1,2}=\frac{7\pm \sqrt{9}}{2}

t1=2t_1=2 and t2=5t_2=5


c) v(0)=602420+60=60v(0)=6\cdot 0^2-42\cdot 0+60=60


d) a(t)=s(t)=v(t)=12t42a(t)=s^{\prime \prime}(t)=v^\prime (t)=12t-42


e) a(0)=12042=42a(0)=12\cdot 0-42=-42


Answer: a) s(3)=45s(3)=45 ; b) t1=2t_1=2 and t2=5t_2=5 ; c) v(0)=60v(0)=60 ; d) a(t)=12t42a(t)=12t-42 ; e) a(0)=42a(0)=-42 .


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