Question

the breaks applied to a car produce an acceleration of 6m/s 2 in the opposite direction to the motion. If the car takes 2s to stop after the application of breaks , calculate the distance it travels during this time.


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Answer 1

Step-by-step explanation:

Given:

Accerleration: a=−6 m/s

2

Time t=2 s

Final velocity, v=0 m/s

v=u+at

0=u−6×2

u=12 m/s

s=ut+

2

1

at

2

s=12×2+

2

1

×(−6)×4

⇒s=12 m

Answer 2

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${ \large \underline{ \red{Question:-}}}$

The breaks applied to a car produce an acceleration of 6m/s 2 in the opposite direction to the motion. If the car takes 2s to stop after the application of breaks , calculate the distance it travels during this time.

Don't spam please

${ \large \underline{ \green{ answer:-}}}$

Given: ✍

Accerleration: a=−6 m/s ^ 2

Time t=2 s

Final velocity, v=0 m/s

v=u+at

0=u−6×2

u=12 m/s

s=ut+1/2 at ^2

s=12×2+ 1 / 2 ×(−6)×4

⇒s=12 m

${ \large \underline{ \orange{ hence \: verified:-}}}$

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${ \large \underline{ \blue{ i \: hope \: it \: is \: helpful \: for \: you:-}}}$