Question #5ef69

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Answer 1 · 2017-06-02T08:10:15

See the answer below

#(v)#

#(dv)/dt# is the acceleration of the kite

#(vi)#
graph{x^3-4x^2+4x [-0.845, 10.25, -1.685, 3.865]}

The gradients at #t=0.7# and #t=2# is #=0#

The acceleration is #=0# at #t=0.7# and #t=2#

This is the local max. and the local min.

#(vii)#

The area under the curve represents the distance travelled.

You can estimate the distance travelled by the kite by counting the number of squares between the curve and the x-axis

#(vii)#

The distance is

#s=intv(t)#dt

#=int_0^4(t^3-4t^2+4t)dt#

#=[t^4/4-4/3t^3+4/2t^2]_0^4#

#=(4*4*4-256/3+32)-(0+0+0)#

#=64+32-256/3#

#=96-256/3#

#=32/3#

#=10.67#

I hope that this is helpful !!