Question #870b3

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Answer 1 · 2017-10-23T15:26:10

The acceleration is #=2(c+6d)ms^-2#

The derivative of #x^n# is

#(x^n)'=nx^(n-1)#

The displacement is

#s(t)=a+bt+ct^2+dt^3#

The velocity is the derivative of the displacement

#v(t)=s'(t)=b+2ct+3dt^2#

The acceleration is the derivative of the velocity

#a(t)=v'(t)=a''(t)=2c+6dt#

When #t=2s#

#a(2)=2c+6d*2=2c+12d=2(c+6d)ms^-2#

Answer 2 · 2017-10-23T15:32:01

Let's have a look.

Acceleration #(a)# is the first order derivative of velocity #(v)#, which in turn is first order derivative of displacement #(S)#.

Therefore, acceleration is the second order derivative of displacement.

Now given that #rarr#

#S=dt^3+ct^2+bt+a#

#:.v=d/dt(S)#

#:.v=3dt^2+2ct+b#

#:.a=d/dt(v)=(d^2S)/(dt^2)#

#:.a=6dt+2c#

Now, here acceleration #(a)# is a function of time #(t)#.

#:.a=f(t)#

#:.a_((t=2))=6dcdot(2)+2c#

#:.a_((t=2))=12d+2c#

#:.a_((t=2))=2(c+6d) ms^(-2)#. (Answer).

Hope it Helps:)