Question

what is the unitial velocity when the distance is 62.5m and retardation is 5m/ s square​

Answer 1

Answer:

Initial velocity of the object is 25 m/s

Step-by-step explanation:

Given:

Distance = 62.5 m  (S)

Acceleration = -5 m/s^2  (a)

Final velocity = 0 m/s  (v)

To find:

Initial velocity   (u)

Using the third equation of motion which says:

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

Substituting the above given values, we get:

$0^{2} -u^{2} =2 \times -5 \times 62.5$

$-u^{2} =2 \times -5 \times 62.5$

$-u^{2}= -10 \times 62.5$

$-u^{2} =-625$

$u^{2}=625$

$u=\sqrt{625}$

$u=25 m/s$

The initial velocity is equal to 25 m/s and then it will reatard at 5 m/s^2 to cover a distance of 62.5m

Answer 2

Given:

• Distance, s = 62.5 m

• Acceleration, a = -5 m/s²

• Final velocity, v = 0 m/s (Rest)

To be calculated:

Calculate the initial velocity ( u ) .

Formula used:

v² = u² + 2as

Solution:

We know that,

v² = u² + 2as

Putting the values in the third equation of motion:

(0)² = u² + 2 × (-5) × 62.5

-u² = -10 × 625/10

-u² = -6250/10

-u² = -625

u² = 625

u = 25 m/s

Thus, the initial velocity of given body is 25 m/s.