Question

Obtain a relation for distance travelled by an object moving with a uniform acceleration in interval between 4th and 5th second.

Answer 1

From the equation of motion

We know,

$s=ut+\frac{1}{2}a{t}^2$

Now distance traveled in 5 sec.

$s=5u+\frac{25}{2}a$

Similarly distance traveled in 4 sec.

$s=4u+\frac{16}{2}a=4u+8a$

We can say that the distance traveled between the 4th and 5th second will be given as

=(Distance traveled in 5 sec) - (Distance traveled in 4 sec)

$\Rightarrow 5u+12.5a-(4u+8a)$

$\Rightarrow 5u+12.5a-4u-8a=u-4.5a$

So the required relation is = u - 4.5 a.