Question
Obtain a relation for the distance travelled by an object moving with a uniform acceleration in the interval between $4th$ and $5th$ seconds.
✓
Answer
From second equation of motion, Distance travelled in t see
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
Distance travelled in $4s$
$\text{s}_4=\text{u}\times4+\frac{1}{2}\text{a}(4)^2$
$=4\text{u}+\frac{1}{2}\times\text{a}\times16=4\text{u}+8\text{a} ( \text{s}_4=$ distance travelled in $4th$ sec$)$
Again,
distance travelled in $5s$
$\text{s}_5=\text{ut}+\frac{1}{2}\text{at}^2$
$=\text{u}\times5+\frac{1}{2}\text{a}(5)^2=5\text{u}+\frac{25}{2}\text{a} ( \text{s}_5=$ distance travelled in $5th$ sec$)$
So, distance travelled in the interval between $4th$ and $5th$ second.
$\text{s}=\text{s}_5-\text{s}_4=\Big(5\text{u}+\frac{25}{2}\text{a}\Big)-4\text{u}+8\text{a}$
$=5\text{u}+\frac{25}{2}\text{a}-4\text{u}-8\text{a}$
$=5\text{u}-4\text{u}\frac{25}{2}\text{a}=8\text{a}$
$=\text{u}+\frac{25\text{a}-16\text{a}}{2}=\text{u}+\frac{9}{2}\text{a}$
so, the relation will be $(\text{u}+\frac{9}{2}\text{a})$
$\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$
Distance travelled in $4s$
$\text{s}_4=\text{u}\times4+\frac{1}{2}\text{a}(4)^2$
$=4\text{u}+\frac{1}{2}\times\text{a}\times16=4\text{u}+8\text{a} ( \text{s}_4=$ distance travelled in $4th$ sec$)$
Again,
distance travelled in $5s$
$\text{s}_5=\text{ut}+\frac{1}{2}\text{at}^2$
$=\text{u}\times5+\frac{1}{2}\text{a}(5)^2=5\text{u}+\frac{25}{2}\text{a} ( \text{s}_5=$ distance travelled in $5th$ sec$)$
So, distance travelled in the interval between $4th$ and $5th$ second.
$\text{s}=\text{s}_5-\text{s}_4=\Big(5\text{u}+\frac{25}{2}\text{a}\Big)-4\text{u}+8\text{a}$
$=5\text{u}+\frac{25}{2}\text{a}-4\text{u}-8\text{a}$
$=5\text{u}-4\text{u}\frac{25}{2}\text{a}=8\text{a}$
$=\text{u}+\frac{25\text{a}-16\text{a}}{2}=\text{u}+\frac{9}{2}\text{a}$
so, the relation will be $(\text{u}+\frac{9}{2}\text{a})$
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