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.
s(t)=2t3−21t2+60ts(t)=2t^3-21t^2+60ts(t)=2t3−21t2+60t
a) s(3)=2⋅33−21⋅32+60⋅3=54−189+180=45s(3)=2\cdot 3^3-21\cdot 3^2+60\cdot 3=54-189+180=45s(3)=2⋅33−21⋅32+60⋅3=54−189+180=45
b) The body is at rest when v(t)=0v(t)=0v(t)=0 .
v(t)=s′(t)=6t2−42t+60=0v(t)=s^\prime (t)=6t^2-42t+60=0v(t)=s′(t)=6t2−42t+60=0
6t2−42t+60=06t^2-42t+60=06t2−42t+60=0
t2−7t+10=0t^2-7t+10=0t2−7t+10=0
D=(−7)2−4⋅10=9D=(-7)^2-4\cdot10=9D=(−7)2−4⋅10=9
t1,2=7±92t_{1,2}=\frac{7\pm \sqrt{9}}{2}t1,2=27±9
t1=2t_1=2t1=2 and t2=5t_2=5t2=5
c) v(0)=6⋅02−42⋅0+60=60v(0)=6\cdot 0^2-42\cdot 0+60=60v(0)=6⋅02−42⋅0+60=60
d) a(t)=s′′(t)=v′(t)=12t−42a(t)=s^{\prime \prime}(t)=v^\prime (t)=12t-42a(t)=s′′(t)=v′(t)=12t−42
e) a(0)=12⋅0−42=−42a(0)=12\cdot 0-42=-42a(0)=12⋅0−42=−42
Answer: a) s(3)=45s(3)=45s(3)=45 ; b) t1=2t_1=2t1=2 and t2=5t_2=5t2=5 ; c) v(0)=60v(0)=60v(0)=60 ; d) a(t)=12t−42a(t)=12t-42a(t)=12t−42 ; e) a(0)=−42a(0)=-42a(0)=−42 .
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