Question

Q5. A body starts to slide over a horizontal surface with an initial
Velocity of 0.5 m/s. Due to Friction, its Velocity decreases at the rate of
0.05 m/s? How much time will it take for the body to Stop?​

Answer 1

Answer:

• 10 sec

Explanation:

As per the information provided in the question, We have :

• A body starts to slide over a horizontal surface with an initial Velocity of 0.5 m/s.
• Due to Friction, its Velocity decreases at the rate of 0.05 m/s.

We are asked to find how much time will it take for the body to stop.

In order to find the time taken, We need to use the first equation of motion.

Here,

• Initial velocity (Given) = 0.5 m/s
• Final velocity = 0 m/s ( As it stops at last )
• Acceleration = 0.05 ( As acceleration is the rate of change in velocity )

First equation of motion is given by –

$\twoheadrightarrow \bf v = u + at$

Putting the given values,

$\twoheadrightarrow \rm 0 = 0.5 + ( - 0.05) \times t$

By simplifying,

$\twoheadrightarrow \rm 0 = 0.5 - 0.05 \times t$

On transposing 0.5 to left hand side,

$\twoheadrightarrow \rm - 0.5 = - 0.05 \times t$

Transposing -0.05 to the left hand side,

$\twoheadrightarrow \rm t = \dfrac{ - 0.5}{ - 0.05}$

By cancelling minus sign and numbers,

$\twoheadrightarrow \rm t =\cancel{ \dfrac{ - 0.5}{ - 0.05} }$

$\twoheadrightarrow \bf t = 10\: sec$

∴ The body will take 10 secs to stop.

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M O R E ⠀TO⠀ KNOW :

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• $\begin{gathered}\sf {First\: equation\: of \: motion} = \bf{v = u + at}\\\end{gathered}$

• $\sf {Second\: equation\: of \: motion} = \bf{v^{2} = u^{2} + 2as}$

• $\sf {Third\: equation\: of \: motion} = \bf{s = ut + \dfrac{1}{2}at^{2}}$

• $\sf {Fourth\: equation\: of \: motion} = \bf{s = vt - \dfrac{1}{2}at^{2}}$

• $\sf {Fifth \: equation\: of \: motion} = \bf{s = \dfrac{1}{2}\Big(u + v\Big)t}$