Question

a skateboarder is traveling at 8m/s. he slows and come to a stop in 4 sec. what was the acceleration? ​

Answer 1

Answer:

• -2 m/s²

Explanation:

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

• A skateboarder is traveling at 8m/s. he slows and come to a stop in 4 sec.

We are asked to find acceleration.

To find acceleration we will use the first equation of motion, i.e (Vf =Vi + at), This can be written as, A = vf - vi / t. Let's put the given values in the equation. The resultant will be the acceleration.

$\longmapsto \bf A = \dfrac{vf - vi }{t}$

Where,

• Vf = Final velocity = 0 m/s
• Vi = Initial velocity = 8 m/s
• T = Time taken = 4 sec

$\longmapsto \rm A = \dfrac{0 - 8}{4}$

$\longmapsto \rm A = \dfrac{-8}{4}$

By cancelling,

$\longmapsto \rm A = \cancel{ \dfrac{-8}{4} }$

$\longmapsto \bf A = \dfrac{-2}{1} \: {m}/{s}^{2}$

∴ Hence, Acceleration is -2 m/s².

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M O R E ⠀TO⠀ KNOW :

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

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

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

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

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