Question

A particle is dropped from a height 80 metre distance travelled by a particle in first second and third second are respectively​

Answer 1

The distance travelled by the particle in the first second and the third second is 5 m and 25 m respectively.

Given,

Height from which the particle was dropped=80 m.

To find,

the distance travelled in the first second and the third second respectively.

Solution:

• Distance is equal to the product of speed and time.
• Distance=Speed x Time.
• As mentioned in the question that the particle is dropped so the initial velocity, u will be 0.
• When a body is in a motion having a constant acceleration, the distance travelled by it in the nth second is given as:
• $s=u+\frac{a}{2} (2n-1)$.
• where, s-distance(or displacement), a-acceleration, n-nth second.
• As this is a case of motion under gravity so the acceleration here will be a=-g.
• $a=-10m/ s^{2}$.

The distance travelled in the first second(n=1) will be:

$s=u+\frac{a}{2} (2n-1)\\s=0+\frac{-10}{2} (2X1-1)\\s=-5(2-1)\\s=-5X1\\s=-5 m.$

The displacement came negative as the body moves downward.

Thus, the distance will be 5 m.

The distance travelled in the third second(n=3) will be:

$s=u+\frac{a}{2} (2n-1)\\s=0+\frac{-10}{2} (2X3-1)\\s=-5(6-1)\\s=-5X5\\s=-25 m.$

The distance travelled is 25 m.

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Answer 2

Answer:

The answer to the given question is the distance travelled by a particle in the first second is -5 m and the third second is -25 m

Explanation:

Given :

A particle is dropped from a height of 80 metres.

To find :

distance travelled by a particle in the first second and third second

Solution :

The formula to find the distance is the product of speed and time.

It is given that the particle is dropped from the height, so the initial velocity will be zero.

The distance travelled by a body at an nth second when the body is in a motion having a constant acceleration.

The formula to find the distance is

$s = u + \frac{a}{2} (2n - 1)$

The u in the formula is the initial velocity, a is the acceleration and n is the nth second, s is the displacement.

The acceleration due to gravity is negative here because the motion is under gravity.

g is negative -g.

The value of g is -10 m/s.

The distance travelled in the first-second n= 1 is substituted n=1.

$s = 0 + \frac{ - 10}{2} (2(1) - 1)$

on solving, we get the answer as

$- 5 (2 - 1) \\ - 5(1) \\ - 5$

The distance travelled in the first second is -5 m

The distance travelled in the third and second is

substitute n=3.

$s = 0 + \frac{ - 10}{2} (2(3) - 1) \\ - 5(6 - 1) \\ - 5(5) \\ - 25$

The distance travelled by a particle in the third second is -25 m.

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