Question

A body moves over horizontal surface with intial velocity of 2m/s velocity decreases uniformly at rate of 0.25m/s^2how much time will it take to stop

Answer 1

Answer:

$\large{\underline{\boxed{\sf Time = 8 \: seconds. }}}$

Explanation:

Given that;

Initial velocity (u) = 2 m/s

Final velocity (v) = 0 m/s

Acceleration (a) = - 0.25 m/s²

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Now, using first equation of motion ;

• v = u + at

⇒ 0 = 2 - (0.25) * t

⇒ 0.25t = 2

⇒ t = $\sf \dfrac{2}{0.25}$

⇒ t = 8

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Hence, the total time taken to stop the body is 8 seconds.

Answer 2

» A body moves over horizontal surface with intial velocity of 2m/s decreases uniformly at rate of 0.25m/s².

• Initial velocity (u) = 2 m/s [starting]

• Final velocity (v) = 0 m/s [because later speed decreases. So, after reaching destination it's velocity is zero]

• Acceleration (a) = - 0.25 m/s²

______________________________

$\boxed{*\:v\:-\:u\:=\:at}$

$\implies$ 0 - 2 = - (0.25)t

$\implies$ - 0.25t = - 2

$\implies$ t = $\dfrac{2}{0.25}$

$\implies$ t = 8 sec.

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$\textbf{The car will stop after 8 sec.}$

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