Question

A ball is initially moving with a velocity 0.8 m/s. Its velocity decreases at a rate of 0.08 m/s². How much time will it take to stop?​

Answer 1

10 seconds

Explanation:

Let the time be t

Velocity given = 0.8 m/s

Retardation (Opposite Acceleration) = 0.08 m/s²

So we will use the formula,

$=> v = u + at$

Solving it:

$v = u + at$

$= > 0 = 0.8 + ( - 0.08 \times t)$

$= >0 = 0.8 - 0.08t$

$= > - 0.08t = - 0.8$

$= > t = \frac{ - 0.8}{ - 0.08}$

$= > t = 10$

So we can conclude that the Time is 10 seconds.

SOME OTHER IMPORTANT FOMULAES:

$v = u + at$

$s = \frac{1}{2} (u + v)t$

or,

$s = ut + \frac{1}{2} a {t}^{2}$

${v}^{2} = {u}^{2} + 2as$

In the above given Formulas,

1. u = Initial Velocity
2. v = Final Velocity
3. a = Acceleration
4. t = Time
5. s = Distance

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