Question

b) A bus was moving with a speed of 54 km/h. On applying brakes it stopped in 8 seconds. Calculate the acceleration.

Answer 1

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Initial Speed  (u)= 54 km /hr = 54 *1000 m / 3600 s = 54*5/18 = 15 m/s

         ( 1km = 1000 m and 1hr = 60 mins = 60*60 Seconds = 3600 seconds)

        Time = 8 seconds

       Final Velocity (V) = 0

 we know that ,

         v = u + at

         0 = 15 + a*8

       0 = 15 + 8a

       a= -15/8 m/s²

   Therefore , acceleration is -15/8 m/s²

also ,

           S = ut+1/2at²

           S = 15*8² + 1/2 *(-15/8)*8²

              =15*64 - 15*4

             = 15(64 - 4)

             =15(60)

             S=900 m

hence Stopping Distance traveled before stopping is 900m .

Answer 2

Answer:

$\bigstar{\bold{Acceleration\:=\:-1.875\:m/s^{2} }}$

Explanation:

$\Large{\underline{\underline{\bf{Given:}}}}$

• Initial speed (u) = 54 km/hr= 15 m/s
• Final speed (v) = 0 m/s
• Time taken = 8 s

$\Large{\underline{\underline{\bf{To\:Find:}}}}$

• Acceleration (a)

$\Large{\underline{\underline{\bf{Solution:}}}}$

→ By the first equation of motion

  v = u + at

→ Substituting the given datas, we get the value of a

  0 = 15 + a×8

  a = -15/8

$\boxed{\bold{Acceleration\:=\:-1.875\:m/s^{2} }}$

→ Here acceleration is negative since it is retardation.

$\Large{\underline{\underline{\bf{Notes:}}}}$

→ Remember to convert all the given datas to the SI units before calculating.

→ The three equations of motion are:

• v = u + at
• s = ut + 1/2 × a × t²
• v² - u² = 2as