Question

a body starting from rest cover distance 0.45 and velocity 300 km per hour it acceleration will be

Answer 1

Answer:

69.44 m/s²

Explanation:

Given :

• Initial velocity = u = 0 m/s

• Distance travelled = s = 0.45 km = 450 metres

• Final velocity = 300 km/hr

To find :

• Acceleration of the body

300 km/hr = 300×5/18 = 250/3 m/s

Using the third equation of motion :

V²-u²=2as

(250/3)²-0²=2×a×450

62500=900a

a = 62500/900

a = 625/9

a = 69.44.. m/s²

The acceleration of the body is equal to 69.44...m/s²

Answer 2

Answer:

Given:

• A body starting from rest cover distance 0.45 and velocity 300 km per hour.

Find:

• What is the acceleration.

Know terms:

1. Initial velocity = (u)
2. Final velocity = (v)
3. Acceleration = (a)
4. Distance covered = (s)

Calculations:

⇒ $\bold{300 = 300 \times \: \dfrac{5}{18}}$

⇒ ${\sf{\underline{\boxed{\red{\sf{ \dfrac{250}{3} }}}}}}$

Using formula:

★ ${\sf{\underline{\boxed{\orange{\sf{v^2 - u^2 =2as }}}}}}$

Calculations

⇒ $\bold{(\dfrac{250}{3})^2 -0^2=2 \times a \times 450}$

⇒ $\bold{62500=900a}$

⇒ $\bold{a = \cancel{\dfrac{62500}{900}}}$

⇒ $\bold{a = \cancel{\dfrac{625}{9}}}$

⇒ ${\sf{\underline{\boxed{\red{\sf{a = 69.44 \: m/s^2}}}}}}$

Therefore, 69.44 m/s² is the acceleration.