Explanation:

_______________________________________________________________________

\text{\huge\underline{\red{Question:-}}}

Question:-

:\Longrightarrow⟹ 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?

\text{\huge\underline{\orange{Given:-}}}

Given:-

:\Longrightarrow⟹ Acceleration , a = -6m/s².

:\Longrightarrow⟹ Time , t = 2s.

:\Longrightarrow⟹ FINAL velocity, v = 0m/s.

\text{\huge\underline{\green{To\ find:-}}}

To find:-

:\Longrightarrow⟹ If the car takes 2s to stop after the application of breaks , calculate the distance it travels during this time

\text{\large\underline{\BLUE{By\ using\ formula:-}}}

By using formula:-

:\Longrightarrow⟹ 1. v = U + at.

:\Longrightarrow⟹ 2. s = u + \mathrm{\dfrac{1}{2}}

2

1

at².

\text{\huge\underline{\purple{Solution:-}}}

Solution:-

:\Longrightarrow⟹ v = u + at

:\Longrightarrow⟹ 0 = u - 6 × 2

:\Longrightarrow⟹ u = 12m/s.

:\Longrightarrow⟹ s = u + \mathrm{\dfrac{1}{2}}

2

1

at²

:\Longrightarrow⟹ s = 12 × 2 + \mathrm{\dfrac{1}{2}}

2

1

× (-6) × 4

:\Longrightarrow⟹ s = 12m.

\text{\huge\underline{\BF{Hence,\ VERIFIED}}}

Hence, Verified