A particle moves along a straight line with velocity given by #v(t) = 7 - (1.01)^(-t^2)# at time #t>=0#. What is the acceleration of the particle at time #t = 3#?

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Answer 1 · 2018-01-26T19:04:58

The acceleration is #=0.055ms^-2#

#"Reminder"#

Let #y=a^(x^2)#

#lny=x^2lna#

#1/ydy/dx=2xlna#

#dy/dx=2xylna=2xa^(x^2)lna#

#(u/v)'=(u'v-uv')/(v^2)#

#v(t)=7-(1.01)^(-t^2)=7-(101/100)^(-t^2)#

#=7-(100/101)^(t^2)#

#=7-100^(t^2)/101^(t^2)#

And

#(100^(t^2))'=2t*100^(t^2)ln100#

#(101^(t^2))'=2t*101^(t^2)ln101#

Therefore,

#a(t)=v'(t)=(-100^(t^2)/101^(t^2))'#

#=-((2t*100^(t^2)ln100)*(101^(t^2))-(2t*101^(t^2)ln101)*(100^(t^2)))/(101^(t^2))^2#

#=(-2tln100*100^(t^2)+2tln101*100^(t^2))/101^(t^2)#

#=(2t(ln101-ln100))/101^(t^2)*100^(t^2)#

The acceleration when #t=3# is

#a(3)=(2*3*(ln101-ln100))/101^(3^2)*100^(3^2)#

#=0.055ms^-2#