1. With the help of a simple case of an object moving with consta… - Vidyadip

Question

With the help of a simple case of an object moving with constant velocity show that the area under velocity$-$time curve represents the displacement over a given time interval.
Establish the relation

$\text{c}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2$ graphically.

A car moving with a speed of $126\ km/ h$ is brought to a stop within a distance of $200m$. Calculate the retardation of the car and the time required to stop it.

✓

Answer

Let a body with constant velocity $v$, between time $t_1$ and $t_2$ as shown in graph.

Area below $=$ area $\text{ABCD} = v(t_2 - t_1)$
Also, the displacement $=$ velocity $\times $ time = $y(t_2 - t_1)$.
Thus proved.

Slope of $v - t$ graph is constant. So there is uniform acceleration. Area below the graph gives the displacement $(x).$

Displacement $=$ Area of trapezium
$=\text{ON}\times\text{OP}+\frac{1}{2}\text{NA}\times\text{MA}$
$=\text{v}_0\times\text{t}+\frac{1}{2}(\text{t})(\text{v}-\text{v}_0)$
Since $\text{a}=\frac{\text{v}-\text{v}_0}{\text{t}}$
We have, $\text{x}=\text{v}_0\text{t}+\frac{1}{2}\text{t at}=\text{v}_0\text{t}+\frac{1}{2}\text{at}^2$

$u 126\ km/hr = 35m/ sec, v = 0, s = 200m.$

using, $\text{v}^2=\text{u}^2+2\text{as},$
$\text{a}=\frac{\text{v}^2-\text{u}^2}{2\text{s}}=\frac{35^2}{2\times200}$
$=-3.00\ \text{ms}^{-2}$
$\text{v}=\text{u}+\text{at}$
$0=35-3\times\text{t}$
$\text{t}=\frac{35}{3}=11.66\text{ sec.}$

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