Question

A Truck at 54km/hr slows down to 36 km/hr in 10s .find the retardation​

Answer 1

Answer:

Retardation of the train will be – 5m/s$^{2}$

Explanation:

Given:

• u = 54 km/hr

• v = 36 km/hr

where u means initial velocity and v means final velocity.

To find:

Retardation.

Solution:

We know that,

$→ \: a = \frac{\triangle \: v}{\triangle \: t}$

Here, it was given that u = 54 km/hr = 15 m/s

v = 36 km/hr = 10 m/s

So, we can say that, u = 15 m/s and v = 10 m/s.

Now substitute the known values in the given formula, we get.

$a = \frac{10 - 15}{10}$

→ a = –5

a = –5m/s$^{2}$

Thus,‐5m/s$^{2}$ is the retardation.

Answer 2

Required Answer:

As per the provided question,we have:

• Intial velocity (u) = 54km/hr

→ 54 × $\dfrac{5}{18}$

→ 3 × 5

→ 15m/s

• Final velocity (v) = 36km/h

→ 36 × $\dfrac{5}{18}$

→ 2 × 5

→ 10 m/s

• Time taken (t) = 10s

We have to find:

• Retardation

Let's solve it:

By the first equation of motion:

• v = u + at

→ 10 = 15 + ( a × 10 )

→ 10 = 15 + 10a

→ 10a = 10 - 15

→ 10a = -5

→ a = $\dfrac{-5}{10}$

→ a = $\dfrac{-1}{2}$m/s²

Therefore, retardation is 1/2m/s².

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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.

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