Question

Question #d6159

Answer 1

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)
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`