Question

if a particle having an initial velocity of 20 metre second power -1 moves with a uniform acceleration of 10 metre second power minus 2 then the distance covered by it when it attains a final velocity of 80 ms power -1 ​

Answer 1

By 3rd equation of motion,

$= >2as = {v}^{2} - {u}^{2}$

$= > 2 \times 10 \times s = {80}^{2} - {20}^{2}$

$= > 20s = 6400 - 400 = 6000$

$= > s = \frac{6000}{20} = 300$

Therefore distance is 300 m

Answer 2

Given:

The initial velocity of the particle = 20 m/s

The uniform acceleration of the particle = 10 m/s²

The final velocity of the particle = 80 m/s

To find:

The distance covered by the particle.

Solution:

We know that one of the laws of motion is:

v²=u² +2aS

Here v is the final velocity of the particle, u is the initial velocity of the particle, a is the acceleration and S is the distance travelled by the particle.

On substituting the values in the given formula, we get:

80²=20² + 2×10×S

6400 = 400 + 20S

6000 = 20S

S = 300 m

Thus, the distance covered by the particle will be 300 m.