Question

A particle starts moving from rest and goes with a constant accleration of 2m/ssq find the distance travelled in 5/2th second

Answer 1

Given:

• A particle starts from rest.
• It has a acceleration of 2m/s².

To Find:

• The distance travelled in 5/2th second.

Concept Used:

• We will use formula to find the distance travelled in n th second.

Answer:

Here its given that a body starts from rest and acquired a acceleration of 2m/s² . Then we have to find the distance travelled in 5/2 th second.

Now , here's a formula to find the distance travelled in n th second.

$\large{\underline{\boxed{\red{\sf{\hookrightarrow S_{n} =u + \dfrac{1}{2}a(2n-1)}}}}}$

where ,

• u is initial velocity.
• n is the nth second .
• a is acceleration

Using this formula ,

$\sf{\implies S_{n} = 0ms^{-1} + \dfrac{1}{2}\times 2ms^{-2} (2\times \dfrac{5}{2}-1)}$

$\sf{\implies S_{n}= 1(5-1)m}$

${\underline{\boxed{\red{\bf{\longmapsto S_{n}= 4m}}}}}$

Hence the distance travelled in 5/2 th second is 4m .

Some more equations of motion:

• v = u + at .
• 2as = v² - u² .
• s = ut + ½at².