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Motion in a Straight Line — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsMotion in a Straight Line5 Marks
Question
Derive the position$-$velocity relation for uniformly accelerated motion from $v-t$ graph. $OR$
  1. Acceleration is defined as the rate of change of velocity. Suppose we call the rate of change of acceleration as $\text{SLAP}$, then what is the unit of $\text{SLAP}$?
  2. A body travels a distance $s_1$ with velocity $v_1$ and distance $s_2$ with velocity $v_2$ in the same direction. Calculate the average velocity of the body.
✓

Answer

Position$-$velocity relation: Consider an object, moving with a uniform acceleration a along a straight line $OX$, with origin at $0$.

Let the object reach at points $A$ and $B$ at instants $t_1$ and $t_2$.
Let $x_1$ and $x_2$ be the displacements of the objects at time $t_1$ and $t_2$ respectively and $v_1$ and $v_2$ be the velocities of the object at positions $A$ and $B$ respectively.
Acceleration of the object $=\frac{\text{Change in velocity}}{\text{time taken}}$
$\text{a}=\frac{\text{v}_2-\text{v}_1}{\text{t}_2-\text{t}_1}$
$\Rightarrow \text{v}_2-\text{v}_1=\text{a}(\text{t}_2-\text{t}_1)$
$\Rightarrow\text{t}_2-\text{t}_1=\frac{(\text{v}_2-\text{v}_1)}{\text{a}}\ \dots(\text{i})$
From position$-$time relation,
we have $\text{x}_2-\text{x}_1=\Big(\frac{\text{v}_1-\text{v}_2}{2}\Big)(\text{t}_2-\text{t}_2)\ \dots(\text{ii})$
Substituting the value of $(t_2 - t_1)$ in the above equation $(ii)$,
we get $\text{x}_2-\text{x}_1=\Big(\frac{\text{v}_2+\text{v}_1}{2}\Big)\Big(\frac{\text{v}_2-\text{v}_1}{\text{a}}\Big)=\frac{\text{v}^2_2-\text{v}^2_1}{2\text{a}}$
$\text{v}^2_2-\text{v}^2_1=2\text{a}(\text{x}_2-\text{x}_1)\ \dots(\text{iii})$
If $u$ and $v$ are the velocities of an object at position $x_0$ and $x$ respectively,
then using $v_1 = u, v_2= v, x_1 = x_0$ and $x_2 = x$ in $(iii)$,
we get $\text{v}^2-\text{u}^2=2\text{a}(\text{x}_0-\text{x})\ \dots(\text{iv})$ If $\text{x}_0-\text{x}=\text{s},$
then $\text{v}^2-\text{u}^2=2\text{as}$
The above equation is the required position velocity relation.Alternate answer
  1. Given $\text{SLAP} =\frac{\text{Change in acceleration}}{\text{time taken}}$
Unit of $\text{SLAP} =\frac{\text{ms}^{-2}}{\text{s}}=\text{ms}^{-3}$
  1. Given: distanve $-\text{s}_1,\text{s}_2$
Velocities $-\text{v}_1,\text{v}_2$
$\because \text{v}=\frac{\text{s}}{\text{t}}$
$\therefore \text{t}_1=\frac{\text{s}_1}{\text{v}_1},\text{t}_2=\frac{\text{s}_2}{\text{v}_2}$
Total time, t $=\text{t}_1+\text{t}_2=\frac{\text{s}_1}{\text{v}_1}+\frac{\text{s}_2}{\text{v}_2}$
Total distance, $\text{s}=\text{s}_1+\text{s}_2$
$\therefore$ Average velocity $(\text{v}_\text{avg})=\frac{\text{s}}{\text{t}}$
$\Rightarrow \text{v}_\text{avg}=\frac{\text{s}_1+\text{s}_2}{\frac{\text{s}_1}{\text{v}_1}+\frac{\text{s}_2}{\text{v}_2}}$

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