Question

a car moving with speed of 54 per hour is braugest to rest with 10 seconds by applying breacks find the acceleration of the car ​

Answer 1

Required Answer:

Given:

• Initial velocity of car (u) = 54km/h = 15m/s

→ $\cancel{54} \times \dfrac{5}{\cancel{18}}$

→ $3 \times 5$

→ 15m/s

• Final velocity (v) = 0m/s [ As it comes to rest.]

• Time taken (t) = 10 seconds

To calculate:

• Acceleration (a)

Calculation:

By using the first equation of motion:

• v = u + at

→ 0 = 15 + ( a × 10 )

→ 0 = 15 + 10a

→ 10a = 0 - 15

→ 10a = -15

→ a = $\dfrac{-15}{10}$

→ a = $\dfrac{-3}{2}$

Therefore, acceleration of the car is -3/2 m/s^2.

[ Note: Negative sign indicates that it is retardation.]

▬▬▬▬▬▬▬▬▬▬▬▬

More information!

Equations of motion:

• v = u + at
• s = ut + ½at²
• v² – u² = 2as

Where,

• ★ v = final velocity = m/s
• ★ u = initial velocity = m/s
• ★ a = acceleration = m/s²

★ s = distance/displacement = m

• ★ t = time = sec

Remember that!

• When a body starts from rest, its initial velocity is 0.
• When a body comes to stop or applies breaks, its final velocity is 0.

▬▬▬▬▬▬▬▬▬▬▬

Answer 2

u = 54 kmph = 54 × 5/18 m/s = 15 m/s

v = 0 m/s {Rest}

t = 10 s

We know,

Acceleration = (Change in velocity)/(Time taken)

Mathematically,

a = (v – u) × t^-1

⇒a = – u × t^-1

⇒a = – 15 m/s × 1/10 s

⇒a = – 1.5 m/s²