Question

a body starts sliding over a horizontal surface with an initial velocity of 0.5m/s . Due to friction, it's velocity decreases at the rate of 0.05m/s×s. how much time will it take to stop.

Answer 1

as we know the formula of acceleration, a = (v-u)/t
therefore, at = v-u
            => t = (v-u)/a

in this case,
v (final velocity)= 0
u (initial velocity)= 0.5 m/s
a (acceleration)= -0.05 m/s²

Therefore t = (0-0.5)/(-0.005) s
            => t = (-0.5)/(-0.05) s
                   = 0.5/0.05 s
                   = 10 s
Therefore the time required for it to stop is 10 seconds. (ANSWER)

P.S.- sorry if it was too long. anyway, the acceleration value is in negative because it is retardation(negative acceleration).

Answer 2

Answer:

the required answer is 10seconds

Explanation:

$\boxed{Given:}$

Initial velocity (u) = 0.5 m/s

Rate of decrease in velocity/Retardation = -0.05 m/s²

Final velocity (v) = 0 m/s (Body stops)

$\boxed{To Find:}$

Time taken by body to stop (t)

$\boxed{Solution:}$

$\sf From \: 1^{st} \: equation \: of \: motion \: we \:$

$\boxed{ \boxed{ \bf{v = u + at}}}$

By substituting values in the equation we get:

$\begin{gathered}\rm \implies 0 = 0.5 + ( - 0.05)t \\ \\ \rm \implies 0.05t = 0.5 \\ \\ \rm \implies t = \dfrac{0.5}{0.05} \\ \\ \rm \implies t = 10 \: s\end{gathered}$

$\therefore \boxed{\mathfrak{Time \ taken \ by \ body \ to \ stop \ (t) = 10 \ s}}$

▶Additional info◀

Sir Isaac Newton has contributed a lot in the field of Science especially Physics and Chemistry. He gave three laws of Newton which we study in the branch of Dynamics.

• Newton's first law of motion.
• Newton's second law of motion.
• Newton's third law of motion.