Question

A motor car tekes 10sec to cover 30mtr and 12 sec to cover 42mtr .find the uniform acceleration of the car and its velocity at the end of 15sec solution

Answer 1

Answer:

0.25 m/s^2      ;      3.75 m/s

Explanation:

In the first 10 second:

   Distance traveled = 30 m

   Time taken = 10 s

 Thus, velocity is ( 30 / 10 ) m/s or 3 m/s.

In the first 12 second:

  Distance traveled = 42 m

  Time taken = 12 s

 Thus, velocity is ( 42 / 12 ) m/s

Time gap between the change of velocities is 12 s - 10 s = 2 s.

Therefore,

      Acceleration is ( chance in velocity ) / ( change in time )

⇒ { (42/12) - 3 } / 2  m/s^2

⇒ { ( 42 - 36 ) / 12 } / 2  m/s^2

⇒ ( 6 / 12 ) / 2  m/s^2

⇒ 6 / 24  m/s^2

⇒ 0.25 m/s^2 = acceleration( uniform ).

     Now, using v = u + at.

⇒ v( velocity till 15 s ) = 0 + ( 0.25 )( 15 )  m/s

⇒ v = 3.75 m/s