# Question #d6159

##### 1 Answer
Jun 12, 2017

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 $t = 2 \mathmr{and} t = 3$. Say $10 \times$. Must have corresponding value of distance to plot enlarged part of graph. Then starting from $t = 2$ and $t = 3 , 2.9 , 2.8 , 2.7$ $\ldots . . , 2.1$ we calculate $\text{rise"/"run}$ 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 $\text{rise"/"run}$ is $- v e$. Matches well with the angle tangents make with $x$-axis, which is $> {90}^{\circ}$ but $< {180}^{\circ} .$

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Alternative method.

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

Let it be

$s \left(t\right) = {s}_{0} + u t + \frac{1}{2} a {t}^{2}$ ......(1)

To ascertain three unknown. We are required to have three equations. Choosing three points where both $s \mathmr{and} t$ are in whole numbers.

1. At $t = 0$, (1) becomes
$s \left(0\right) = {s}_{0} + u \times 0 + \frac{1}{2} a \times {0}^{2}$
$\implies s \left(0\right) = {s}_{0}$
From the graph $s \left(0\right) = 25 m$. Hence,
${s}_{0} = 25 m$ .......(2)
2. At $t = 4$, (1) becomes
$s \left(4\right) = 25 + u \times 4 + \frac{1}{2} a \times {4}^{2}$
Again from the given graph
$21 = 25 + u \times 4 + \frac{1}{2} a \times {4}^{2}$
$\implies 4 u + 8 a = - 4$
$\implies a = - 0.5 \left(1 + u\right)$ .......(3)
1. At $t = 6$, (1) becomes
$s \left(6\right) = 25 + u \times 6 + \frac{1}{2} a \times {6}^{2}$
Again from the given graph
$16 = 25 + 6 u + 18 a$
$\implies 6 u + 18 a = - 9$
$\implies 2 u + 9 a = - 3$
Using (3) we get
$2 u + 6 \left(- 0.5 \left(1 + u\right)\right) = - 3$
$\implies 2 u - 3 u = - 3 + 3$
$\implies u = 0 m {s}^{-} 1$ .......(4)
Inserting this value in (3) we get
$a = - 0.5 m {s}^{-} 2$ .........(5)

Using (2), (4) and (5) equation (1) becomes
$s \left(t\right) = 25 - 0.25 {t}^{2}$ .....(6)
Graph can be represented as this equation.

We know that velocity $v = \frac{\mathrm{ds}}{\mathrm{dt}}$
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

$v = u + a t$

Using (4) and (5), above equation becomes
$v \left(t\right) = - 0.5 t$ and therefore
$v \left(2\right) = - 0.5 \times 2$
$v \left(2\right) = - 1 m {s}^{-} 1$

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If you like you may verify equation (6) for other points on the graph. For example
$s \left(10\right) = 25 - 0.25 \times {10}^{2}$
$\implies s \left(10\right) = 0 m$