Question

Car movies at apeed of 40km it ia stopped by applying brakes with produces uniform acceleration of _o.6m/s2 how much distance before coming to stop

Answer 1

Answer:

102.9 metres

Explanation:

Given :

• Initial velocity = u = 40 km/h = 40×5/18 = 100/9 m/s

• Acceleration = a = - 0.6 m/s²

• Final velocity = v = 0 m/s

To find :

• Distance covered by the car before stopping

Using the third equation of motion :

• V²-u²=2as

0²-(100/9)²=2×-0.6×s

0-10000/81=-1.2s

-10,000/81= - 1.2s

s = - 10,000/81×-10/12

s = - 1,00,000/972

s=102.9 metres

The distance travelled by the car is equal to 102.9 metres

Answer 2

Answer:

Given:

• A carr moves at speed of 40km and it is stopped by applying brakes with produces uniform acceleration of -0.6 m/s².

Find:

• Distance covered by the car before it stops.

Know terms:

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

Using formula:

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

Note:

• Initial velocity is always 0.
• 40 km/hr is equal to 100/9 m/s.

Calculations:

$\longrightarrow\rm{(0)^{2} - (\dfrac{100}{9})^{2} = 2 \times (-0.6) \times s}$

$\longrightarrow\rm{s = \dfrac{10000}{(81 \times 1.2)}}$

$\longrightarrow{\rm{\underline{\boxed{\red{\rm{102.9 \: m}}}}}}$

Therefore, 102.9 m is the distance covered by the car before it stops.