Question #089b1

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Question #089b1

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Answer 1 · 2016-12-22T19:29:51

Kinematic equation of interest is

$v (t) = u + a t$ $v(t)=u+at$ .....(1)
where $v (t)$ $v(t)$ is velocity after time $t$ $t$ , $u$ $u$ is initial velocity of an object and $a$ $a$ is constant acceleration experienced by it.

Recall the expression
$Displacement = Velocity \times time$ $"Displacement"="Velocity"xx"time"$
Observe it looks like equation of a straight line in the form
$y = m x + c$ $y=mx+c$ .

![revisionworld.com]( useruploads.socratic.org )

We know that velocity is rate of change of displacement, therefore equation (1) can be written as

$\frac{d s (t)}{d t} = u + a t$ $(ds(t))/(dt)=u+at$
$\Rightarrow d s (t) = (u + a t) \cdot d t$ $=>ds(t)=(u+at)cdot dt$ .....(2)

If we integrate both sides we get
$\int d s (t) = \int_{t_{0}}^{t} (u + a t) \cdot d t$ $intds(t)=int_(t_0)^t (u+at)cdot dt$
$\Rightarrow s (t) = \int_{t_{0}}^{t} (u + a t) \cdot d t$ $=>s(t)=int_(t_0)^t (u+at)cdot dt$ ......(3)
We see that LHS of the equation is total displacement, and RHS is area under the velocity-time graph from time $t_{0}$ $t_0$ to $t$ $t$ .
Equation (3) is the required expression.

One should not be surprised if one calculates integral of RHS of equation (3) from time $t = 0$ $t=0$ to $t$ $t$ , one actually obtains the other kinematic equation

$s = u t + \frac{1}{2} a t^{2}$ $s=ut+1/2at^2$

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Question #089b1

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