What is its velocity? What is its acceleration?

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3 Answers
Feb 10, 2018

#x(2.5)=3.435#
#v(2.5)=-1.91#
#a(t)=-2#

Explanation:

It is given that the displacement #x# at time #t# is given by #x(t)=1.96+3.09t-t^2#.

At #t=2.5#, the displacement #x(2.5)=1.96+3.09*2.5-2.5^2=3.435#.

The velocity #v# is the derivative of displacement #x# with respect to time #t#, or #v(t)=dx/dt=3.09-2t#. Thus, the velocity at time #t=2.5# is #v(2.5)=3.09-2*2.5=-1.91#.

The acceleration #a# is the derivative of velocity #v# with respect to time #t#, or #a(t)=(dv)/dt=-2#. Since it stays constant regardless of the time, the acceleration at time #t=2.5# is #-2#.

Feb 10, 2018

The particle moves according to the equation, #1.96 + 3.09t - t^2#
so,
#x=1.96 + 3.09t - t^2#
Velocity is the rate of change of displacement, #x#
therefore, #v=dx/dt#

#v=(d(1.96 + 3.09t - t^2))/dt#

#v=(3.09-2t)m/s#

Now, acceleration is the rate of change of velocity, #v#
therefore, #a=dv/dt#

#a=(d(3.09-2t))/dt#

#a=-2# #m/s^2#

Feb 10, 2018

Given, #x=1.96+3.09t-t^2#

at, #t=2.50# ,#s=1.96+(3.09*2.50)-(2.50)^2=3.435 m#

differentiating the equation w.r.t time,we get,

velocity =#(dx)/(dt) = 3.09 -2t#

So,at #t=2.50s# its velocity is #3.09 -(2*2.50)=-1.91 m/s#

And,differentiating it one again we get, acceleration= #(dv)/(dt) = -2 m/s^2#

ALTERNATIVELY

You can compare the equation with, #s=ut -1/2 at^2#

So,you can rearrange as, #x-1.96=3.09t - 1/2 ×2×t^2#

So comparing we can say,it had an initial velocity of #3.09 m/s#(at #t=0#) acceleration is #-2 m/s^2# and at #t=0# it was #1.96 m# to the left of origin i.e along negative #X# axis.