The position of an object moving along a line is given by #p(t) = 2t^3 - 2t +2#. What is the speed of the object at #t = 4 #?

Question

1488 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-18T13:27:21

#94ms^(-1)#

#p(t)=2t^3-2t+2#

to find the speed we differentiate

#p'(t)=6t^2-2#

for #t=2#

#p'(4)=6xx4^2-2#

speed#=94ms^(-1)#

SI units assumed

Answer 2 · 2017-12-18T13:30:22

The speed is #=94ms^-1#

The speed of an object is the derivative of the position.

#v(t)=(dp)/(dt)#

The position is

#p(t)=2t^3-2t+2#

The speed is

#v(t)=p'(t)=6t^2-2#

And when #t=4#

#v(4)=6*(4)^2-2=96-2=94ms^-1#