Question #d6159

1 Answer
Jun 12, 2017

Answer:

There is no way one can find the exact value of velocity at #t=2# without drawing the tangent line at the desired point.

Explanation:

At the best one can find an approximate value of velocity at #t=2# by enlarging the portion of graph between #t=2 and t=3#. Say #10xx#. Must have corresponding value of distance to plot enlarged part of graph. Then starting from #t=2# and #t=3,2.9,2.8,2.7# #.....,2.1# we calculate #"rise"/"run"# for each interval. As #t# approaches #t=2,# (for #t=2.1#) we get a good approximate value of velocity at #t=2#.
Notice that value obtained for #"rise"/"run"# is #-ve#. Matches well with the angle tangents make with #x#-axis, which is #>90^@# but #<180^@.#

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Alternative method.

We need to find kinematic expression represented by the given graph.

Let it be

#s(t)=s_0+ut+1/2at^2# ......(1)

To ascertain three unknown. We are required to have three equations. Choosing three points where both #sand t# are in whole numbers.

  1. At #t=0#, (1) becomes
    #s(0)=s_0+uxx0+1/2axx0^2#
    #=>s(0)=s_0#
    From the graph #s(0)=25m#. Hence,
    #s_0=25m# .......(2)
  2. At #t=4#, (1) becomes
    #s(4)=25+uxx4+1/2axx4^2#
    Again from the given graph
    #21=25+uxx4+1/2axx4^2#
    #=>4u+8a=-4#
    #=>a=-0.5(1+u)# .......(3)
    1. At #t=6#, (1) becomes
      #s(6)=25+uxx6+1/2axx6^2#
      Again from the given graph
      #16=25+6u+18a#
      #=>6u+18a=-9#
      #=>2u+9a=-3#
      Using (3) we get
      #2u+6(-0.5(1+u))=-3#
      #=>2u-3u=-3+3#
      #=>u=0ms^-1# .......(4)
      Inserting this value in (3) we get
      #a=-0.5ms^-2# .........(5)

Using (2), (4) and (5) equation (1) becomes
#s(t)=25-0.25t^2# .....(6)
Graph can be represented as this equation.

We know that velocity #v=(ds)/dt#
To find out velocity, either we differentiate (6) with respect to time #t# and evaluate the expression at #t=2#.
Or we may use the following kinematic expression

#v=u+at#

Using (4) and (5), above equation becomes
#v(t)=-0.5t# and therefore
#v(2)=-0.5xx2#
#v(2)=-1ms^-1#

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If you like you may verify equation (6) for other points on the graph. For example
#s(10)=25-0.25xx10^2#
#=>s(10)=0m#