Question

a body starting with initial velocity u moves with constant acceleration a find the expression for distance travelled in nth seconds

Answer 1

Answer:

Refer to the pic above.

Thankyou

Answer 2

Answer:

Explanation:

Disclaimer:

A body starting with initial velocity u moves with a constant accelearation a. Find the expression for distance travelled in nth second. A body was dropped from a very high building, calculate its distance travelled in $5^{th}$ second.

Concept:

The concept of motion in physics will be used to solve this question.

Given:

The initial velocity of the body is $u$ which is moving with constant acceleration.

To Find:

We need to find the expression for distance which was travelled in nth second and the value of the distance in $5^{th}$ seconds.

Solution:

Here $S_{n}$ is the distance .

The formula of $S_{n}$ is (Distance travelled in $t$ seconds - Distance travelled in $(t-1)$ seconds.

Then Distance travelled in $t$ seconds = $[ut+\frac{at^{2} }{2} ]$

Distance travelled in $(t-1)$ seconds = $[u(t-1)+\frac{a(t-1)^{2} }{2} ]$

We can write the expression as $S_{n}=[ut+\frac{at^{2} }{2} ]-[u(t-1)+\frac{a(t-1)^{2} }{2} ]$

By solving this equation we will get $u+\frac{a(2 n-1)}{2}$

Therefore $u+\frac{a(2 n-1)}{2}=u+a\left(n-\frac{1}{2}\right)$

Put the value in the above equation $S_{n} =0+15(5-\frac{1}{2} )=45 m$ .

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