Question #b49a3

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Answer 1 · 2017-06-01T10:29:37

See the answer below

The formula is

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

$(i)$ ,

After $2s$

$v(2)=2^3-4*2^2+8=0ms^-1$

$(ii)$ ,

The derivative is

$(dv)/dt=v'(t)=3t^2-8t+4$

$(iii)$

$(dv)/dt=0$

When,

$3t^2-8t+4=0$

$(3t-2)(t-2)=0$ , $=>$

$t=2/3$ , and $t=2$

$(iv)$

Therefore

$v(2/3)=(2/3)^3-4*(2/3)^2+4*2/3=1.19ms^-1$

$v(2)=2^3-4*2^2+4*2=0$

The local maximum value is $=1.19ms^-1$ and the minimum value is $=0ms^1$

Here is a graph
graph{x^3-4x^2+4x [0, 7.024, 0, 3.51]}

Hope that will be helpful