Question

a body starts from raste and moves with uniform accelaration of 4m/s2 how much Time Will it take to cover a distance of 128m​

Answer 1

Answer:

Explanation:

Initial velocity = 0 m/2

Acceleration = 4 m/s^2

distance = 128 cm = 0.128 m

using the equation d = Vi * t + (1/2) *a*t^2

since Vi = 0 , the term Vi * t = 0

therefore:

d = (1/2)at^2

rearranged for time:

square root of (2d)/a = square root of [(0.256m)/(4m/s^2)]

= sqrt 0.64

= 0.8 seconds

Answer 2

Given :-

➜ A body starts from rest, So the Initial velocity, u = 0 m/s

➜ Moves with a uniform acceleration, a = 4 m/s²

To Find :-

➜ Time taken by the body to cover a distance of 128 m

Solution :-

Since, The acceleration is uniform throughout the condition.

∴ Using the second Equation of motion.

⇒ h = ut + 1/2 at²

⇒ 128 = 0×t + (4 × t²)/2

⇒ 128 = 2t²

⇒ t² = 64

⇒ t = 8, -8

Because Time can't be negative,

∴ Time taken, t = 8 seconds

Some Information :-

☛ Three equations of motion:

1. v = u + at
2. h = ut + 1/2 at²
3. 2as = v² - u²

Where,

• v = Final Velocity
• u = Initial velocity
• a = Acceleration
• h = Height reached ( or Distance travelled )
• s = Distance travelled
• t = Time taken