$(a)$ Derive second equation of motion $S=u t+\frac{1}{2} a t^{2}$ graphically where the symbols have their usual meanings.
$(b)$ A car accelerates uniformly from $18\, km h ^{-1}$ to $36\, km h^{-1}$ in $5$ seconds. Calculate the acceleration and the distance covered by the car in that time.
$(a)$ Derive second equation of motion $S=u t+\frac{1}{2} a t^{2}$ graphically where the symbols have their usual meanings.
$(b)$ A car accelerates uniformly from $18\, km h ^{-1}$ to $36\, km h^{-1}$ in $5$ seconds. Calculate the acceleration and the distance covered by the car in that time.
$(a)$ The area under the graph is the area of the rectangle $OACD$ plus the area of the triangle $ABC$ on top of it as shown in figure. The rectangle has a height $u$ and a length $t$. This area is the distance travelled by the object.
Hence, $S=u t+\frac{1}{2} \times t \times(v-u)$ $....(1)$
But from the expression $v=u+a t,$ we have
at $=v-u$. Substituting in equation $(1),$ we have
$S=u t+\frac{1}{2}(a t) t=u t+\frac{1}{2} a t^{2}$ $...(2)$
$(b)$ Given $u=18 km h ^{-1}=5 m s ^{-1}$
$v=36 km h ^{-1}=10 m s ^{-1}, t=5 s$
Using $v=u+a t,$ we have
$a=v-\frac{u}{t}=\frac{(10-5)}{5}=1 m s ^{-2}$
Also,
$S=u t+\frac{1}{2} a t^{2}=5 \times 5+\frac{1}{2} \times 1 \times(5)^{2}$
$=37.5 m$
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