Question

a man starting from a rest moving with uniform acceleration 5m/s^2 for 4 sec.what is his final velocity and distance covered?​

Answer 1

Given that, a man starts from rest (means the initial velocity of the man i.e. u is 0 m/s).

Further given that, a man starting from a rest moving with uniform acceleration 5m/s² for 4 sec.

Here, acceleration (a) = 5 m/s² and time (t) = 4 sec

We have to find the final velocity (v) of the man and the distance (s) covered by the man.

a) For final velocity:

Using the First Equation of Motion;

v = u + at

→ v = 0 + (5)(4)

→ v = 20

Therefore, the final velocity of the man is 20 m/s.

b) For distance:

Using the Third Equation of Motion,

v² - u² = 2as

→ (20)² - (0)² = 2 × 5 × s

→ 400 - 0 = 10s

→ 10s = 400

→ s = 40

Therefore, the distance covered by the man is 40 m.

Answer 2

Given :-

A man starting from a rest moving with uniform acceleration 5 m/s² for 4 sec.

• Acceleration(a) = 5 m/s²

• Time(t) = 4 s

To find :-

The final velocity(v) and the distance covered(s).

Solution :-

Initial velocity(u) = 0 m/s      ...(As the man starts from rest)

We know,

Final velocity = Sum of Initial velocity and (product of Acceleration and Time)

v = u + at

Putting the known values in the equation :-

v = 0 + (5*4)

⇒v = 0 + 20

⇒v = 20 m/s

We also know,

sq. of final velocity + sq. of initial velocity = product of acceleration and distance

v² - u² = 2 as

Putting the known values in the equation :-

(20)² - (0)² = 2*5*s

⇒400 - 0 = 10s

⇒10s = 400

⇒s = 400/10

⇒s = 40 m

∴ So, the final velocity of the man is 20 m/s and the man covered 40 m.