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