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

AcademicPhysicsNCERTClass 9

Let $u\ m/s$ be the initial velocity of a moving body with uniform acceleration $a$.

We know, the second equation of motion,

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

Here, $u\rightarrow$ initial velocity

$t\rightarrow$ time

$a\rightarrow$ acceleration

$s\rightarrow$distance travelled

So, distance travelled in $4th$ second $s_{4th}=u\times4+\frac{1}{2}a\times4^2$

$s_{4th}=4u+8a$

Similarly, $s_{5th}=u\times5+\frac{1}{2}a\times5^2$

Or $s_{5th}=5u+\frac{25}{2}a$

So, the distance traveled between the interval of $4^{th}$ and $5^{th}$ seconds

$=s_{5th}-s_{4th}$

$=(5u+\frac{25}{2}a)-(4u+8a)$

$=u+\frac{9}{2}a$

raja
Updated on 10-Oct-2022 13:28:47

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