Question

A body starts from rest and attains a acceleration of 0.5 m/s^2. calculate the velocity at the end of 3 minutes. also, find the distance travelled by it during that time.

Answer 1

Answer:

Explanation:

Solution,

Here, we have

Initial velocity, u = 0 m/s (As starts from rest)

Acceleration, a = 0.5 m/s².

Time taken, t = 3 min = 3 × 60 = 180 seconds

To Find,

Final velocity, v and,

Distance covered, s

Formula to be used,

1st and 3rd equation of motion,

v = u + at and v² - u² = 2as

So, putting all the values, we get

v = u + at

⇒ v = 0 + 0.5 × 180

⇒ v = 0.5 × 180

⇒ v = 90 m/s.

Hence, the final velocity is 90 m/s.

Now, Distance covered,

v² - u² = 2as

⇒ (90)² - (0)² = 2 × 0.5 × s

⇒ 8100 = 1s

⇒ s = 8100 m

Hence3, the distance covered is 8100 m.

Answer 2

Answer

Velocity at the end of 3 min = 90 m/s

Distance traveled by body during this 3 min = 8100 m

Given

A body starts from rest and attains a acceleration of 0.5 m/s²

To Find

Velocity at the end of 3 minutes

Distance travelled by body during this time

Concept Used

As the acceleration is constant throughout the motion , so we need to apply equation's of kinematics .

→ v = u + at

→ s = ut + ¹/₂ at²

→ v² - u² = 2as

Solution

Initial velocity , u = 0 m/s

[ ∵ body starts from rest ]

Acceleration , a = 0.5 m/s²

Final velocity , v = ? m/s

Time , t = 3 min = 3 × 60 s = 180 s

Distance , s = ? m

Apply 1st equation of motion ,

⇒ v = u + at

⇒ v = (0) + (0.5)(180)

⇒ v = 0 + 90

⇒ v = 90 m/s

So , Final velocity = 90 m/s

________________________

Apply 3rd equation of motion ,

⇒ v² - u² = 2as

⇒ (90)² - (0)² = 2(0.5)s

⇒ 8100 - 0 = s

⇒ 8100 = s

⇒ s = 8100 m

So , Distance travelled = 8100 m