Question

A car moving with a speed of 54km/hr on apply; breaks stopped in 8sec . Calculate the acceleration?

Answer 1

For a body moving with initial velocity u, accelerates with Acceleration a, covers a distance s in time t, the final velocity v.

then,

Equations of motion are,

$\boxed{ \: \: \: \: \: v = u + at \: \: } \\ \boxed{s = ut + \frac{1}{2} at ^{2} } \\ \boxed{ {v}^{2} - {u}^{2} = 2as}$

Solution :

Initial velocity = 54 km/hr = 54 * 5/18 m/s = 15m/s

Time taken to stop = 8 seconds.

Final velocity = 0 m/s ( As the vehicle stopped)

We have,

$v = u \: + at \: \\ \\ a = \frac{v - u}{t} \\ \\ a = \frac{0 - 15}{8} \\ \\ a = \frac{ - 15}{8} = - 1.875m {s}^{ - 2}$

Therefore, The Acceleration of the vehicle is - 1.875 m/s²

Answer 2

$\huge{\mathfrak{\underline{\underline{\pink{AnSwEr:-}}}}}$

Acceleration = 1.875

$\LARGE{\mathfrak{\underline{\underline{\blue{Explanation:-}}}}}$

________________________

Given :-

As we know that object becomes at rest

So ,

Final Velocity (v) = 0 km/hr

Initial Velocity (u) = 54 km/hr

Time (t) = 8 sec

______________________

To Find :-

Acceleration

______________________

Solution :-

We have to convert Initial Velocity into m/s

So, Multiply it by 5/18

⇒ u = 54 × 5/18

⇒ u = 15 m/s

We have equation :-

$\LARGE{\bf{\underline{\boxed{v \: = \: u \: + \: at}}}}$

__________________[Put Values]

⇒ 0 = 15 + a(8)

⇒ 18a = -15

⇒ a = -15/8

⇒ a = - 1.875

⇒ a ≈ - 1.9 m/s²

Negative Sign of acceleration means that it is applied in opposite direction.

$\large{\underline{\boxed{\red{a \: = \: 1.9 m\s^{2}}}}}$

$\rule{200}{2}$

♡ Remember ♡

• v = u + at
• S = ut + 1/2 × at²
• v² - u² = 2as