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Question

A particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in tth and (t+1)th seconds is 100cm, then its velocity after t seconds, in cm/s, is


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Solution

Step 1. Given data

The sum of the distance travelled in tth and (t+1)th seconds is 100cm.

We have to find the velocity.

Step 2. Concept used.

The distance travelled in nth second is,

Sn=u+122n-1a ----------- 1

Here, S is the distance travelled time, u is the initial velocity, and a is acceleration of the body in motion.

Step 3. Calculate the velocity.

The distance travelled in tth and (t+1)th second is 100cm.

Now we will find the sum of distances.

By using the expression 1, the distance travelled in tth is,

St=u+122t-1a ------ 2

And the distance travelled in (t+1)th is,

St+1=u+122t+1-1a

=u+122t+1a---------- 3

According to the given, the total distance travelled in tth and (t+1)th second is 100cm, so, we have to add expression 1 and 2, we get,

St+St+1=100

Substitute the values of expression 1 and 2 in above expression, we get,

u+122t-1a+u+122t+1a=100

2u+at=100

u+at=1002

u+at=50 ------4

Now, from the first equation of motion the velocity of any particle after time t, if it moves with acceleration a is,

v=u+at -------- 5

Now, its clear that u+at is similar in expression 5 as well as 4, so substitute the value of u+at from expression 4 in 5, we get

v=u+at

v=50cm/s

Hence, a particle is moving in a straight line with initial velocity and uniform acceleration a. If the sum of the distance travelled in tth and (t+1)th seconds is 100cm, then its velocity after t seconds, in cm/s, is 50cm/s.


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