Obtain a relation for the distance travelled by an object moving …

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 $(\text{u}+\frac{9}{2}\text{a})$

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.